# Telescoping Series - Product

A **telescoping series of product** is a series where each term can be represented in a certain form, such that the multiplication of all of the terms results in massive cancellation of numerators and denominators. This process is similar to telescoping sum, in which we have massive cancellation of addition in one term with subtraction in the subsequent term.

The simplest form of a telescoping product occurs when $a_k = \dfrac{ t_k} { t_{k+1}}$, in which case the product of these $n$ terms is equal to

$a_ 1 \times a_2 \times \cdots \times a_n = \frac{t_1}{t_2} \times \frac{t_2}{t_3} \times \cdots \times \frac{ t_n}{t_{n+1} } = \frac{ t_1 } { t_{n+1} }.$

Observe that the cancellation of each denominator with the numerator of the subsequent term allows us to arrive at the final result immediately.

## Basic Form

**Explanation of the product notation $\prod$**:

Before we get started with the calculation, it would be better to simplify the expression into a simple form so that it is easier for the readers to digest the information, than to write it in the long form $a_1 \times a_2 \times \cdots \times a_n$. This is because it will become an inconvenience to convey the same message especially when a shorter description is always preferred. We will be introducing a symbol $\prod$ to describe the product of such terms.

Recall that the summation notation $\sum$ provides for us a condensed way to express patterns in sum. Well, the product notation $\prod$ is not that different, the only difference being that we use it to express patterns in product instead of sum. For example, with the factorials notation, we can write $n!$ as

$n! = 1 \times2\times 3 \times \cdots \times (n-1) \times n = \prod_{k=1}^n k.$

Likewise, we can also write the product of the first 10 positive odd numbers as $\displaystyle \prod_{k=1}^{10} (2k-1)$.

Now that we have established all the preliminaries, let's get started on the interesting part: telescoping product!

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**Simple example to see**:

Before we jump into the massive cancellation part of the calculation, let us begin with an introductory example of how telescoping product can be motivated. Consider the product

$\prod_{k=1}^{9} \dfrac k{k+1} = \dfrac 12 \times \dfrac23 \times \cdots \times \dfrac9{10} .$

It might be daunting at first to find the value of the multiplication of so many fractions, right? Don't be afraid! Let's try to find the product of first few terms one at a time:

$\begin{array}{ l c l } \dfrac{1}2 &=& \dfrac12 \\\\ \dfrac12 \times \dfrac23 &=& \dfrac13 \\\\ \dfrac12 \times \dfrac23 \times \dfrac34 &=& \dfrac14 \\\\ \dfrac12 \times \dfrac23 \times \dfrac34 \times \dfrac45 &=& \dfrac15. \end{array}$

All the 4 equations represent the product of the first $n$ fractions in the product $\prod_{k=1}^{9} \frac k{k+1}$. Did you notice that all the final values of these equations are in the form of $\frac 1{n+1}$, where $n=1,2,3,4?$ Is this true for $n=9$ as well? Let's find out!

Recall that we are essentially multiplying fractions, so if the numerator of one of the fraction is equal to the denominator of another fraction (both of which are non-zero), we can cancel them off in pairs and thus have

$\dfrac1{\cancel2} \times \dfrac{\cancel 2}3 = \dfrac13 .$

Similarly, cancelling off the 2's and 3's in pairs shows that $\dfrac1{\cancel2} \times \dfrac{\cancel2}{\cancel3}\times \dfrac{\cancel3}4 = \dfrac14$. Analogously, we can compute the product in question by cancelling off almost all the numbers in pairs:

$\prod_{k=1}^{9} \dfrac k{k+1} = \dfrac 1{\cancel2} \times \dfrac {\cancel2}{\cancel3} \times \dfrac {\cancel3}{\cancel4}\times \dfrac {\cancel4}{\cancel5} \times \dfrac {\cancel5}{\cancel6} \times \dfrac {\cancel6}{\cancel7} \times \dfrac {\cancel7}{\cancel8} \times \dfrac {\cancel8}{\cancel9}\times \dfrac {\cancel9}{10} =\dfrac1{10} .$

Let's try out one more example that applies this same concept.

We have finally computed the multiplication of so many fractions by telescoping! Now, let's read the following section where we discuss more examples on how to form a telescoping product.

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**Explanation of how $\frac{T(n\pm k)}{T(n)}$ works**:

In the previous sections, we have computed a product of fractions by cancelling the terms in pairs, and each of these pairs consists of one term from the numerator of a fraction and the other from the denominator of another fraction. One natural question arises: how do we systematically identify which terms to cancel out with which? For example, by writing out the first few terms of the product below, we still are not able to cancel out any terms, right?

$\prod_{m=1}^{10} \dfrac m{m+5} = \dfrac1{6} \times \dfrac2{7} \times \dfrac{3}{8} \times \cdots \times \cdots \times \dfrac{10}{15}$

Well, the technique is actually the same, but we need a slight adjustment to show that there exist such terms that are being cancelled off, as shown below:

$\begin{aligned} \prod_{m=1}^{10} \dfrac m{m+5} &=& \dfrac1{\cancel6} \times \dfrac2{\cancel7} \times \dfrac{3}{\cancel8} \times \dfrac4{\cancel9} \times \dfrac5{\cancel{10}} \times \dfrac{\cancel6}{11} \times \dfrac{\cancel7}{12} \times \dfrac{\cancel8}{13} \times \dfrac{\cancel9}{14} \times \dfrac{\cancel{10}}{15} \\ &=& \dfrac{1\times2\times3\times4\times5}{11\times12\times13\times14\times15} = \dfrac1{3003} . \end{aligned}$

However this is rather cumbersome and tedious, isn't it? Are we supposed to write out all the fractions to identify which terms should be cancelled off? No, we are not.

Notice that for each of the fractions in the product $\prod_{m=1}^{10} \frac m{m+5}$, the numerator is always 5 less than the denominator, so only after the $5^\text{th}$ term are we able to see that the terms start to cancel out. So we can rewrite this as $\prod_{m=1}^{10} \frac{T(m-5)}{T(m)}$, where $T(m) = m+5$. With the fraction (in the product) written as a function of $m$, it will be easier to determine how a product telescopes.

To generalize all such telescoping products (products that are able to telescope, i.e. able to cancel out 1 or more terms in pairs), we define $\displaystyle \prod_{m} S(m)$ to be able to telescope if it can be expressed in the form of $\prod \frac{T(m\pm k)}{T(m) }$ for some integer $k$. To illustrate this point, let us define two terms: backward cancellation and forward cancellation.

Backward cancellationis the cancellation of terms in pairs in the case that the term(s) in the denominator of a fraction is(are) cancelled out with the term(s) in the numerator of a later fraction, where the product can be expressed as $\prod \frac{T(m- k)}{T(m) }$ for some positive integer $k$.

Forward cancellationis the cancellation of terms in pairs in the case that the term(s) in the denominator of a fraction is(are) cancelled out with the term(s) in the numerator of a prior fraction, where the product can be expressed as $\prod \frac{T(m+k)}{T(m) }$ for some positive integer $k$.

As an example, the telescoping product on the left below represents a backward cancellation, whereas the telescoping product on the right below represents a forward cancellation:

$\dfrac1{\cancel2} \times \dfrac{\cancel2}{\cancel3}\times \dfrac{\cancel3}4 = \dfrac14 , \qquad \dfrac{\bcancel2}1 \times \dfrac{\bcancel3}{\bcancel2}\times \dfrac4{\bcancel3} = 4.$

Since we are always cancelling terms off from one direction only (from left to right, or from right to left) but never both ways, we call these product forms as **unidirectional forms**.

Let's try out a few examples to make sure we understood these concepts.

Evaluate the product $\prod_{m=1}^{9} \frac{3m-1}{3m+11}$.

As a first step, let us find out whether this product has a backward or a forward cancellation. This allows us to figure out or verify which terms to cancel out first, whether it is in the numerator or denominator.

Let $T(m) =3m - 1$, then $T(m+4) = 3(m+4)-1=3m+11$, so we can express the product as $\prod_{m=1}^9 \frac{T(m)}{T(m+4)}$. Since the product can be expressed in the form of $\prod \frac{T(m- 4)}{T(m) }$, the product telescopes and the cancellation is a backward cancellation.

Now, let's find out from which terms we should start cancelling off terms, and with that information, let's find out how many terms remain after the cancellation.

Since it is in the form of $\prod \frac{T(m- 4)}{T(m) }$, only after the $4^\text{th}$ term do we start cancelling off terms. So we will be cancelling off the $5^\text{th}, 6^\text{th}, 7^\text{th},8^\text{th}$ and $9^\text{th}$ numerators with the $1^\text{st}, 2^\text{nd}, 3^\text{rd}, 4^\text{th}$ and $5^\text{th}$ denominators, which is a total of 5 pairs of terms cancelled off.

So finally, we cancelled off as many terms as we can already, and what's left to do is to evaluate this expression. From the above, the product of the remaining terms that are not cancelled is

$\dfrac{(3\cdot 1 - 1)(3\cdot 2- 1)(3\cdot 3 - 1)(3\cdot 4 - 1)}{(3\cdot 6 +11)(3\cdot 7 +11)(3\cdot 8 +11)(3\cdot 9 +11)} = \dfrac{11}{15428} .\ _\square$

For each of the examples provided above, we have used only one of backward and forward cancellations, and thus they are strictly in unidirectional form. But is it possible to have a telescoping product that utilizes both these cancellations at the same time? Let's find that out in the next section.

## Advanced Form

In the previous section, we learned two similar unidirectional cancellations in a telescoping product: backward and forward cancellations. The next thing we should ask ourselves is, "What if there's a combination of these unidirectional cancellations? Is it still possible for the terms to cancel out in pairs?" Yes! We just need to identify and partition the expression into products of strictly backward and strictly forward cancellations. We call such expressions as *bidirectional forms*.

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**Combination of 2 unidirectional forms**

We shall begin with a combination of 2 unidirectional forms by diving straight into an example:

$\prod_{r=3}^{100} \dfrac{r^2-1}{r^2-4}.$

The fraction in the product looks horrendous, doesn't it? Fear not! This fraction can be factorized. Note that both the expressions in the numerator $\left(r^2-1\right)$ and denominator $\left(r^2-4\right)$ are in the form of a difference of two squares identity: $a^2-b^2=(a+b)(a-b) .$ Thus, we have

$\prod_{r=3}^{100} \dfrac{r^2-1}{r^2-4} = \prod_{r=3}^{100} \dfrac{\color{green}{(r-1)} \color{purple}{(r+1)}}{\color{green}{(r-2)}\color{purple}{(r+2)}}.$

Because $\prod f$ and $\prod g$ each have a finite non-zero value, we can use the property $\prod (f \cdot g) = \prod f \cdot \prod g$ to get

$\prod_{r=3}^{100} \dfrac{r^2-1}{r^2-4} = \prod_{r=3}^{100} \dfrac{r-1}{r-2} \times \prod_{r=3}^{100} \dfrac{r+1}{r+2} .$

What remains is to evaluate these two products and multiply them! For the first product,

$\prod_{r=3}^{100} \dfrac{r-1}{r-2} = \dfrac21 \times \dfrac32 \times \dfrac43 \times \dfrac54 \times \cdots \times \dfrac{98}{97} \times \dfrac{99}{98}.$

Now this looks familiar, doesn't it? Yes it does! It's a forward cancellation, because $\frac{r-1}{r-2}$ is in the form of $\frac{S(r)}{S(r-k)}$ for a positive integer $k$. We can now cancel out in pairs:

$\begin{aligned} \prod_{r=3}^{100} \dfrac{r-1}{r-2} & = \dfrac21 \times \dfrac32 \times \dfrac43 \times \dfrac54 \times \cdots \times \dfrac{98}{97} \times \dfrac{99}{98} \\ &= \dfrac{\cancel2}{1} \times \dfrac{\cancel3}{\cancel2} \times \dfrac{\cancel4}{\cancel3} \times \dfrac{\cancel5}{\cancel4} \times \cdots \times \dfrac{\cancel{98}}{\cancel{97}} \times \dfrac{99}{\cancel{98}} \\ &= \dfrac{99}1 = 99. \end{aligned}$

Analogously, we can see that the second product is a backward cancellation:

$\begin{aligned} \prod_{r=3}^{100} \dfrac{r+1}{r+2} & = \dfrac{4}{\bcancel5} \times \dfrac{\bcancel5}{\bcancel6} \times \dfrac{\bcancel6}{\bcancel7} \times \cdots \times \dfrac{\bcancel{100}}{\bcancel{101}} \times \dfrac{\bcancel{101}}{102} \\ & = \dfrac4{102} = \dfrac2{51}. \end{aligned}$

Putting them all together yields

$\prod_{r=3}^{100} \dfrac{r^2-1}{r^2-4} = \prod_{r=3}^{100} \dfrac{r-1}{r-2} \times \prod_{r=3}^{100} \dfrac{r+1}{r+2} = 99 \times \dfrac2{51} = \dfrac{66}{17},$

and you're done!

Food for thought: Can we still evaluate the same product above if we split the product $\prod_{r=3}^{100} \frac{r^2-1}{r^2-4}$ as the multiplication of $\prod_{r=3}^{100} \frac{r-1}{r+2}$ and $\prod_{r=3}^{100} \frac{r+1}{r-2}$ instead?

As you can see in the illustration above, the product represents a product of more than one unidirectional forms, so we call such products as bidirectional forms.

Using the concepts that we've just learned, can you solve the following problems?

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**Combination of more than 2 unidirectional forms**

Now that you're capable of handling a combination of 2 unidirectional forms, is it possible to push further? That is, is it possible to find the combination of more than 2 unidirectional forms? The answer is still yes! Let's use a modified version of the very first product we used earlier in this section, $\prod_{n=3}^{20} \frac{(n^2-1)(n^2-4)}{n^4}$.

Applying the difference of two squares identity, we have $n^2 - 1 = (n-1)(n+1)$ and $n^2 - 4 = (n-2)(n+2)$, so

$\dfrac{(n^2-1)(n^2-4)}{n^4} = \dfrac{(n-1)(n+1)(n-2)(n+2)}{n^4} = \dfrac {n-1}n \times \dfrac{n+1}n \times \dfrac{n+2}n \times \dfrac{n+2} n.$

Now, we're back to the combination of backward and foward cancellations again:

$\begin{aligned} \prod_{n=3}^{20} \dfrac{(n^2-1)(n^2-4)}{n^4} &= \prod_{n=3}^{20} \dfrac{(n-1)(n+1)(n-2)(n+2)}{n^4} \\ &= \prod_{n=3}^{20} \dfrac{n-1}n \times \prod_{n=3}^{20} \dfrac{n+1}n \times \prod_{n=3}^{20} \dfrac{n-2}n\times \prod_{n=3}^{20} \dfrac{n+2}n. \end{aligned}$

Then we just need to evaluate the 2 backward cancellations and 2 forward cancellations. Using the concepts we learned in the unidirectional forms, we have

$\prod_{n=3}^{20} \dfrac{n-1}n = \dfrac{2}{20}, \quad \prod_{n=3}^{20} \dfrac{n+1}n = \dfrac{21}{3}, \quad \prod_{n=3}^{20} \dfrac{n-2}n = \dfrac{1\times2}{19\times20}, \quad \prod_{n=3}^{20} \dfrac{n+2}n = \dfrac{21\times 22}{3\times4}.$

Calculating the product of these four values gives the desired answer of $\frac{539}{3800}$, and we're done again.

As you can see in the above illustration, technically there's no difference in techniques whether it be 2, or more than 2 combinations of unidirectional forms.

Now let's try out the following example that uses this very same idea:

## Determining How a Product Telescopes

We have learned about the different forms which allow for a telescoping product. This is easy to determine if everything is presented clearly, as shown above. However, since that is rarely going to be the case, this section deals with determining if a given product allows for a telescoping form.

To get started, test small cases. If the final result looks simple or seems to have a general form to it, then it provides some evidence that there is cancelling via a telescoping product approach. It remains to find suitable sequences of $\frac{T(n+k)}{T(n)}$ or $\frac{S(n-k)}{S(n)}$. We present 2 approaches of figuring out these terms.

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**Approach 1: factoring of terms**

- Factorize the general term as far as possible: suppose that $a_k = \frac{ b_k \times c_k}{d_k \times e_k }$.
- List out terms of the form $b_{k-2}, b_{k-1}, b_{k}, b_{k+1}, b_{k+2}$, and compare them with terms $d_{k-2}, d_{k-1}, d_{k}, d_{k+1}, d_{k+2}$ and likewise for $c_k$ and $e_k$.
- If two of these terms are equal, then these sequences will pair up and result in cancellation.

Evaluate

$M_k = \prod_{k=2}^{n} \frac{k^3-1}{k^3+1}.$

Start by testing small cases:

- $n = 2: M_2 = \frac{7 }{ 9}$
- $n = 3: M_3 = \frac{7}{9} \times \frac{ 26}{28} = \frac{ 13}{18}$
- $n = 4: M_4 = \frac{13}{18} \times \frac{ 63}{65} = \frac{7}{10} .$
Now, despite having denominators of 9, 28, 65, we see that the final term is extremely small, suggesting that we can cancel these terms. Since we have an algebraic expression, let's factorize the general term:

$a_k = \frac{ (k-1)(k^2+k+1) }{(k+1) (k^2 - k + 1 ) }.$

Let's label $b_k = k-1, c_k = k^2 + k + 1, d_k = k+1, e_k = k^2 - k + 1$. Then it should be clear how $b_k, d_k$ cancel out, but let's follow the procedure listed above:

$\begin{array} { l | l l } n & b_n & d_n \\ \hline k-2 & k-3 & k-1 \\ k-1 & k-2 & k \\ k & k-1 & k+ 1 \\ k+1 & k & k+2 \\ k+2 & k+1 & k+3 \\ \end{array}$

Now, it is apparent that $b_k = d_{k-2},$ which allows the canceling.

Let's proceed similarly with $c_k, e_k$. Right now, it is not clear if, or how, they will cancel out, since they look so different. Once again, we list out the terms:

$\begin{array} { l | l l } n & c_n & e_n \\ \hline k-2 & k^2 - 3k + 1 & k^2 - 5k + 3 \\ k-1 & k^2 - k + 1 & k^2 - 3k + 1 \\ k & k^2 + k + 1 & k^2 - k + 1 \\ k+1 & k^2 + 3k + 3 & k^2 + k + 1 \\ k+2 & k^2 + 5k + 7 & k^2 + 3k + 3 \\ \end{array}$

Now, it is apparent that $c_k = e_{k+1}$, which allows the canceling.

We now write $a_k = \frac{ b_k } { b_{k+2} } \times \frac{ c_k } { c_{k-1}},$ which allows us to telescope and obtain

$\prod_{k=2} ^ {n} a_k = \prod_{k=2}^n \frac{b_k} { b_{k+2}} \times \prod_{k=2} ^ n \frac{ c_k } { c_{k-1} } = \frac { b_2 b_3} { b_{k+1} b_{k+2}} \times \frac{ c_n} { c_1 } = \frac{ 2 (n^2 + n + 1 ) } { 3 n(n+1) }.\ _\square$

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**Approach 2: non-obvious factoring**

- The general term is a single term, because $\frac{T(n) } { T(n+1)}$ simplifies.
- It is most often explained as "multiply and divide by the same term."
- We have to make a guess at what $T(n)$ is, guided by various identities that we are familiar with.

Evaluate $(1+x)\big(1+x^2\big)\big(1+x^4\big) \big(1+x^8\big) \big( 1 + x^ {16} \big)$.

Of course, one could expand out the $2^ 5$ terms to see what we get. In the spirit of this section, let's consider instead how we could write this as a telescoping product.

Using the identity that $\frac{ a^2 - b^2 }{ a-b } = a+b$, we get

$\begin{aligned} 1+ x & = \frac{ 1- x^2 } { 1-x } \\\\ 1+ x^2& = \frac{ 1- x^4 } { 1-x^2 } \\\\ 1+ x^4 & = \frac{ 1- x^8 } { 1-x^4 } \\\\ 1+ x^8 & = \frac{ 1- x^{16} } { 1-x^8 } \\\\ 1+ x^{16} & = \frac{ 1- x^{32} } { 1-x^{16} }. \end{aligned}$

We thus get a forward cancellation, and the product is equal to $\dfrac{ 1 - x^{32} } { 1 - x}$. $_\square$

Evaluate $\cos \theta \times \cos 2 \theta \times \cos 4 \theta \times \cos 8 \theta \times \cos 16 \theta$.

In this case, it's harder to guess how to proceed. If we tried expressing $\cos n \theta$ in terms of $\cos \theta$, we will end up with a mess, and it's not clear why we would have a nice answer.

We use the identity $\frac{ \sin 2 \theta } { \sin \theta } = 2 \cos \theta$. Then we get

$\begin{aligned} 2 \cos \theta & = \frac{ \sin 2 \theta } { \sin \theta } \\\\ 2 \cos 2 \theta & = \frac{ \sin 4 \theta } { \sin 2 \theta } \\\\ 2 \cos 4 \theta & = \frac{ \sin 8 \theta } { \sin 4 \theta } \\\\ 2 \cos 8 \theta & = \frac{ \sin 16 \theta } { \sin 8 \theta } \\\\ 2 \cos 16 \theta & = \frac{ \sin 32 \theta } { \sin 16 \theta }. \end{aligned}$

Thus, the product can be written as $\dfrac{ \sin 32 \theta } { 32 \sin \theta }$. $_\square$

Now, let's work on the following problems to practice identifying these forms:

Evaluate

$\frac {(8^4 + 64) ( 16^4 + 64)(24^4+64)(32^4+64)(40^4+64)(48^4 + 64)(56^4+64)(64^4+64) } { (4^4+64)(12^4+64)(20^4+64)(28^4+64)(36^4+64)(44^4+64)(52^4+64)(60^4+64) }.$

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**Details and Assumptions:**

- Make sure you scroll right (if necessary) to see the full fraction. The dot on the extreme right is a full stop (punctuation mark).

Two infinite products $A$ and $B$ are defined as follows:

$A = \displaystyle\prod\limits_{n=2}^{\infty} \left(1-\frac{1}{n^3}\right), \quad B =\displaystyle\prod\limits_{n=1}^{\infty}\left(1+\frac{1}{n(n+1)}\right).$

If $\frac{A}{B} = \frac{m}{k},$ where $m$ and $k$ are relatively prime positive integers, determine $100m+k$.

## Common Pitfalls

In this section, we will be discussing common mistakes students make when solving telescoping products, as it is very easy for students to commit such errors, especially when dealing with infinities.

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**(1) Not remembering what the last term is, especially for an infinite product**:

Often times, when we cancel off terms, we forget what terms actually remain. For example, the following calculation for the infinite product is wrong:

$\begin{aligned} \displaystyle \prod_{r=1}^\infty \dfrac r{r+1} &=& \dfrac12 \times \dfrac23 \times \dfrac34 \times \dfrac45 \times \cdots \\ &=& \dfrac1{\cancel2} \times \dfrac{\cancel2}{\cancel3}\times \dfrac{\cancel3}{\cancel4}\times \dfrac{\cancel4}{\cancel5} \times \cdots \\ &=& 1. \end{aligned}$

Weird, isn't it? The product of fractions that are all smaller than 1 produces a number that is equal to 1? How is that possible? Can you spot the error?

The mistake here is that we didn't consider its partial product and that's why we think that every term except the first number, 1, is being canceled out in pairs, whereas in actuality there should be more than one term remaining. By calculating the infinite product, we are in fact taking the limit of its partial product:

$\begin{aligned} \displaystyle \prod_{r=1}^\infty \dfrac r{r+1} &=& \displaystyle \lim_{R\to\infty}\prod_{r=1}^R\dfrac r{r+1} \\ &=&\lim_{R\to\infty} \left( \dfrac12 \times \dfrac23 \times \dfrac34 \times \dfrac45 \times \cdots \times \dfrac{R-1}R \times \dfrac R{R+1} \right) \\ &=& \lim_{R\to\infty} \left( \dfrac1{\cancel2} \times \dfrac{\cancel2}{\cancel3}\times \dfrac{\cancel3}{\cancel4}\times \dfrac{\cancel4}{\cancel5} \times \cdots \times \dfrac{\cancel{R-1}}{\cancel R} \times \dfrac{\cancel R}{R+1} \right) \\ &=& \lim_{R\to\infty} \left( 1 \times \dfrac1{R+1} \right) = 1\times0 = 0. \end{aligned}$

Let us try a slight variation on the product above.

Find the error in the following working:

$\begin{aligned} \prod_{r=1}^\infty \dfrac r{r+3}& =& \dfrac14 \times\dfrac25 \times \dfrac36 \times \dfrac47 \times \dfrac5{8} \times\dfrac69 \times \dfrac7{10}\times \cdots \\ & =&\dfrac1{\cancel4} \times\dfrac2{\cancel5} \times \dfrac3{\cancel6} \times \dfrac4{\cancel7} \times \dfrac{\cancel5}{\cancel8} \times\dfrac{\cancel6}{\cancel9} \times \dfrac{\cancel7}{\cancel{10}}\times \cdots \\ &=& 1. \end{aligned}$

As before, it appears that we have found that the product of fractions that are smaller than 1 produces a number that is equal to 1. The mistake made here (that most students make) is that they

cancel off the terms that they can see. That is, they knew that, for each new fraction they write up, all the numerators and denominators would be pairwise cancelled off. But what they didn't account for is that that they didn't treat it as a partial product; that is, it is not always true that the terms remaining are the first few terms that were left uncancelled.Here's the correct working (by partial product):

$\begin{aligned} \prod_{r=1}^\infty \dfrac r{r+3}& =& \lim_{R\to\infty} \prod_{r=1}^R\dfrac r{r+3} \\ & =& \lim_{R\to\infty} \left( \dfrac14 \times\dfrac25 \times \dfrac36 \times \dfrac47 \times \dfrac5{8} \times\dfrac69 \times \dfrac7{10}\times \cdots \times \dfrac{R-2}{R+1}\times \dfrac{R-1}{R+2} \times \dfrac{R}{R+3} \right) \\ & =& \lim_{R\to\infty} \left( \dfrac1{\cancel4} \times\dfrac2{\cancel5} \times \dfrac3{\cancel6} \times \dfrac{\cancel4}{\cancel7} \times \dfrac{\cancel5}{\cancel8} \times\dfrac{\cancel6}{\cancel9} \times \dfrac{\cancel7}{\cancel{10}}\times \cdots \times \dfrac{\cancel{R-2}}{R+1}\times \dfrac{R-1}{R+2} \times \dfrac{R}{R+3} \right) \\ & =& \lim_{R\to\infty} \dfrac{1\times2\times3}{(R+1)(R+2)(R+3) } = 0.\ _\square \end{aligned}$

Now, should we find the partial product for the product in the following example as well? Why or why not?

$$

**(2) Attempting to cancel far-away terms**:

Notice that for all the backward cancellation and forward cancellation that occur, the distance between each pair of terms is always fixed. A common mistake that students make when evaluating telescoping products is that they cancel off terms in pairs whose distances are arbitrary. For example, they may cancel off the denominators of the fraction in the $n^\text{th}$ term with the numerator of the fraction in the $(2n)^\text{th}$ term. So as $n$ becomes larger and larger, the distance between the cancellation of terms becomes unboundedly large. To illustrate this point, let us consider the Wallis product

$\displaystyle \prod_{n=1}^\infty \dfrac{2n\cdot 2n}{(2n-1)(2n+1)} = \dfrac21 \cdot \dfrac23 \cdot \dfrac43\cdot \dfrac45\cdot \dfrac65\cdot \dfrac67\cdot \dfrac87\cdot \dfrac89 \cdot \dfrac{10}9 \cdot \dfrac{10}{11} \cdots = \dfrac\pi2.$

Suppose we start cancelling off the denominators with values of 3 versus the numerators with values of $3\times 2 = 6$, then we get

$\dfrac21 \cdot \dfrac2{\cancel3} \cdot \dfrac4{\cancel3}\cdot \dfrac45\cdot \dfrac{\cancel{6}\ 2}5\cdot \dfrac{\cancel{6}\ 2}7\cdot \dfrac87\cdot \dfrac89 \cdot \dfrac{10}9 \cdot \dfrac{10}{11} \cdots.$

Likewise, suppose we also canceled off the denominators with values of 5 versus the numerators with values of $5\times 2 = 10$, then we get

$\dfrac21 \cdot \dfrac2{\cancel3} \cdot \dfrac4{\cancel3}\cdot \dfrac4{\cancel5}\cdot \dfrac{\cancel{6}\ 2}{\cancel5}\cdot \dfrac{\cancel{6}\ 2}7\cdot \dfrac87\cdot \dfrac89 \cdot \dfrac{\cancel{10}\ 2}9 \cdot \dfrac{\cancel{10}\ 2}{11} \cdots.$

By repeating this process indefinitely, we will be cancelling off the denominators with values of $n$ versus the numerators with values of $2n$, and the resultant number is

$\dfrac21 \cdot \dfrac2{\cancel3} \cdot \dfrac4{\cancel3}\cdot \dfrac4{\cancel5}\cdot \dfrac{\cancel{6}\ 2}{\cancel5}\cdot \dfrac{\cancel{6}\ 2}{\cancel7}\cdot \dfrac8{\cancel7}\cdot \dfrac8{\cancel9} \cdot \dfrac{\cancel{10}\ 2}{\cancel9} \cdot \dfrac{\cancel{10}\ 2}{\cancel{11}} \cdots = \infty,$

because when the fraction is simplified, the numerator consists of the product of infinitely many powers of 2, whereas the denominator is equal to 1 only. And we will make the absurd conclusion that $\frac\pi2 = 1$.

The reason why we have committed a fallacy here is because we make the wrongful assumption that we think it's alright to cancel off terms that have no fixed distance. Another way to phrase it is this: we make the assumption that the logarithm of this telescoping product (which is a telescoping sum) is absolutely convergent when in fact it's not the case here.

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**(3) Breaking up an infinite telescoping product into multiple infinite telescoping products**:

Another common mistake students make when attempting to simplify a telescoping product is that they make the assumption that the product can be expressed as a product of multiple products in attempts to calculate each of these simplified expressions in partition in hopes of making their work easier. To illustrate this point, consider the telescoping product $\prod_{n=3}^\infty \frac{n^2-1}{n^2}$, whose value is equal to $\frac23$ (we will leave it as a exercise for the reader). However, suppose we break the product as such:

$\begin{aligned} \prod_{n=3}^\infty \dfrac{n^2-1}{n^2} &=& \left(\prod_{n=3}^\infty \dfrac{n-1}{n}\right) \times \left(\prod_{n=3}^\infty \dfrac{n+1}{n}\right) \\ &=& \lim_{R\to\infty}\left( \dfrac23 \times \dfrac34 \times \dfrac45 \times \cdots \right) \left( \dfrac43 \times \dfrac54 \times \dfrac65 \times \cdots \right) \\ &=& 0 \times \infty \; , \end{aligned}$

which is an indeterminate form, and is a wrong value because the product holds a finite value.

The proper way to evaluate this product is by using the *combination of more than 2 unidirectional forms* in the section above, or alternatively by finding each of its partial products first before combining the two products, as shown below:

$\begin{aligned} \prod_{n=3}^\infty \dfrac{n^2-1}{n^2} &=& \lim_{R\to\infty} \left(\prod_{n=3}^R \dfrac{n-1}{n}\right) \times \left(\prod_{n=3}^R\dfrac{n+1}{n}\right) \\ &=& \lim_{R\to\infty}\left( \dfrac23 \times \dfrac34 \times \dfrac45 \times \cdots \times \dfrac {R-1}R\right) \left( \dfrac43 \times \dfrac54 \times \dfrac65 \times \cdots\times \dfrac{R+1}R \right) \\ &=& \lim_{R\to\infty} \dfrac2R \times \dfrac {R+1}{3} = \dfrac23. \end{aligned}$

To summarize, calculating telescoping products that deals with infinities can be tricky, and we should always be cautious of their partial products first.

Now that you know how easy it is to make a mistake when dealing with infinities, let's try out a simple problem to ensure we won't make the same mistake twice!

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**(4) Forgetting that the terms have shifted**:

Another common pitfall students make is to make the assumption that all terms can be rearranged. For example, in a previous example, we have shown that $\prod_{r=1}^\infty \frac r{r+3} = 0$. Similarly, we will know that $\frac16 \prod_{r=1}^\infty \frac r{r+3} = 0 \times\frac16 = 0,$ right? Then what is wrong with the following:

$\begin{aligned} \dfrac16\prod_{r=1}^\infty \dfrac r{r+3} &=& \dfrac1{1\times2\times3} \left( \dfrac14 \times \dfrac25 \times \dfrac36 \times \dfrac47 \times \dfrac58 \times \cdots \right) \\ &=& \dfrac1{\cancel1\times2\times3} \left( \dfrac{\cancel1}4 \times \dfrac25 \times \dfrac36 \times \dfrac47 \times \dfrac58 \times \cdots \right) \\ &=& \dfrac1{\cancel1\times\cancel2\times\cancel3} \left( \dfrac{\cancel1}4 \times \dfrac{\cancel2}5 \times \dfrac36 \times \dfrac47 \times \dfrac58 \times \cdots \right) \\ &=& \dfrac1{\cancel1\times\cancel2\times\cancel3} \left( \dfrac{\cancel1}4 \times \dfrac{\cancel2}5 \times \dfrac{\cancel3}6 \times \dfrac47 \times \dfrac58 \times \cdots \right) \\ &=& \dfrac1{\cancel1\times\cancel2\times\cancel3} \left( \dfrac{\cancel1}{\cancel4} \times \dfrac{\cancel2}{\cancel5} \times \dfrac{\cancel3}{\cancel6} \times \dfrac{\cancel4}{\cancel7} \times \dfrac{\cancel5}{\cancel8} \times \cdots \right) \\&=& 1? \end{aligned}$

The terms that are being cancelled out in the bracket is a backward cancellation and the cancellations are all of a fixed distance. Then why does the product (whose value is equal to 0) is equal to 1? The flaw behind this is that we have forgotten about writing it in terms of a its partial product again! That's all! In fact, the right working should consist of the final lines

$\cdots = \lim_{R\to\infty} \dfrac1{\cancel1\times\cancel2\times\cancel3} \left( \dfrac{\cancel1}{\cancel4} \times \dfrac{\cancel2}{\cancel5} \times \dfrac{\cancel3}{\cancel6} \times \dfrac{\cancel4}{\cancel7} \times \dfrac{\cancel5}{\cancel8} \times \cdots \times \dfrac {\cancel{R-2}}{R+1}\times \dfrac {\cancel{R-1}}{R+2}\times \dfrac {\cancel R}{R+3}\right) = 0.$