# Telescoping Series - Sum

A **telescoping series** is a series where each term \( u_k \) can be written as \( u_k = t_{k} - t_{k+1} \) for some series \( t_{k} \). This is a challenging sub-section of algebra that requires the solver to look for patterns in a series of fractions and use lots of logical thinking. These patterns will more than often cause mass cancellation, making the problem solvable by hand. Some patterns are harder to find than others. Often, partial fractions are used here in a way which shall be demonstrated later.

The benefit of such a series is that it allows us to easily add up the terms, because

\[\begin{align} u_1 + u_2 + u_3 + \cdots + u_n &= (t_1 - t_2) + (t_2 - t_3) + (t_3 - t_4 ) + \cdots + (t_n - t_{n+1} ) \\ &= t_1 - t_ {n+1}. \end{align}\]

Observe that most of the \(t_k\) terms cancel out with their counterparts in other parentheses, and thus we are only left with \( t_1 - t_{n+1} \). This is comparable to a collapsible telescope, in which the long spyglass is easily retracted into a small instrument that fits into your pocket.

As you work through Arron's telescoping series investigation, you would realize that for the series \( u_k = \frac{1}{k(k+1)} \) and terms \( t_k = \frac{ 1}{k} \), we have \( u_k = t_k - t_{k+1} \) since

\[ \frac{ 1} { k} - \frac{ 1} { k+1} = \frac{ k+1} { k(k+1) } - \frac{ k} { k(k+1) } = \frac{ 1}{ k (k+1) } . \]

As such, we have our telescoping series. This allows us to conclude that

\[ \sum_{i=1}^n \frac{1}{i(i+1) } = \frac{1}{1} - \frac{ 1}{n+1}. \]

In particular, since \( \frac{1}{n+1} \) approaches 0 as \(n\) gets large, we get that

\[ \sum_{i=1}^\infty \frac{1}{i(i+1) } = 1. \]

Evaluate \(\displaystyle \frac{1}{4} + \frac{1}{28} + \frac{1}{70} + \cdots + \frac{1}{9700}\).

Observe that we can write this expression as \(\frac{1}{1 \times 4} + \frac{1}{ 4 \times 7 } + \frac{1}{ 7 \times 10 } + \cdots + \frac{1}{ 97 \times 100 }\).

Then there is a way to make most part of this cancel itself out, which is

\[\begin{align} \dfrac{1}{1 \times 4} + \dfrac{1}{ 4 \times 7 } + \dfrac{1}{ 7 \times 10 } + \cdots + \dfrac{1}{ 97 \times 100 } &=\dfrac{1}{3} \left(\dfrac{ 4 - 1}{ 1 \times 4 } + \dfrac{ 7 - 4 }{ 4 \times 7 } + \dfrac{ 10 - 7 }{ 7 \times 10 } + \cdots+ \dfrac{ 100 - 97 }{ 97 \times 100 }\right)\\ &=\dfrac{1}{3} \left(\dfrac{1}{1} -\dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{7} + \cdots + \dfrac{1}{97}-\dfrac{1}{100}\right)\\ &=\dfrac{1}{3} \left(1- \dfrac{1}{100}\right)\\ &=\dfrac{1}{3} \left(\dfrac{99}{100}\right)\\ &=\dfrac{33}{100}. \ _\square \end{align}\]

Evaluate \( \displaystyle \frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+ \sqrt{3}} + \frac{1}{\sqrt{3}+ \sqrt{4}} +\cdots+ \frac{1}{\sqrt{99} + \sqrt{100}} . \)

Multiply both the numerator and denominator of each fraction by the conjugate of the denominator. Then, for example, the first term simplifies to

\[ \dfrac{1}{\sqrt{1} + \sqrt{2}} \cdot \dfrac{\sqrt{2} - \sqrt{1}}{\sqrt{2} - \sqrt{1}} = \sqrt{2} - \sqrt{1}. \]

If you keep doing this for each successive fraction, you'll keep getting \( 1 \) in the denominator to obtain

\[ (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \cdots. \]

Like a telescope, it all can be collapsed and the last fraction simplifies to \( \sqrt{100} - \sqrt{99} \). Since it's the \( \sqrt{99} \) that will be canceled out, the given expression is equal to

\[ \sqrt{100} - \sqrt{1} = 10-1 = 9. \ _\square \]

Evaluate \( \displaystyle \sum_{n=1}^{2015} \dfrac{1}{n^2 + 3n + 2} \).

For this type of problem, you need some knowledge of partial fractions.

Rewrite \( \frac{1}{n^2 + 3n + 2} \) in the form \( \frac{a}{n+1} + \frac{b}{n+2} \). Then we can say that \( a(n+2) + b(n+1) = 1. \)

An easy way to figure out the values of \(a\) and \(b\) is to plug in some easy numbers. For example, let \( n= -2\). Then we get \( 0a - b = 1 \), so \( b = -1\). If we let \( n = -1 \), then we get \( a + 0b = 1 \), so \( a = 1 \). Our partial fraction decomposition has yielded that \( \frac{1}{n^2 + 3n + 2} = \frac{1}{n+1} - \frac{1}{n+2} \).

Now, we have \( \displaystyle \sum_{n = 1}^{2015} \left[ \dfrac{1}{n+1} - \dfrac{1}{n+2} \right]\). This means we have

\[ \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} +\cdots+ \dfrac{1}{2015} - \dfrac{1}{2016} + \dfrac{1}{2016} - \dfrac{1}{2017}. \]

Notice how a bunch of terms can be canceled out, like a collapsing telescope.

This leaves

\[ \dfrac{1}{2} - \dfrac{1}{2017} = \dfrac{2015}{4034}. \ _\square \]

Let \(F_n\) be the Fibonacci sequence with \(F_0 = F_1 = 1\) and \(F_{n+2}=F_{n+1}+F_n\) for all \(n\geq0\).

Prove that \(\displaystyle \sum_{k=1}^{n} \dfrac{1}{F_{k-1}F_{k+1}}<1\) for all \(n\geq 1\).

For all \(k\geq 1\), we have

\[\begin{align} \dfrac{1}{F_{k-1}F_{k+1}}&=\dfrac{F_k}{F_{k-1}F_kF_{k+1}}\\ &=\dfrac{F_{k+1}-F_{k-1}}{F_{k-1}F_kF_{k+1}}\\ &=\dfrac{1}{F_{k-1}F_k}-\dfrac{1}{F_kF_{k+1}}. \end{align}\]

As such, we have our telescoping series. This allows us to conclude that

\[\sum_{k=1}^{n} \dfrac{1}{F_{k-1}F_{k+1}}=\dfrac{1}{F_0F_1}-\dfrac{1}{F_nF_{n+1}}=1-\dfrac{1}{F_nF_{n+1}}<1. \ _\square\]

\[\begin{align} a_n&=1+6(n-1) \\ b_n&=1+21(n-1) \\ c_n&=202+102(n-1) \end{align} \]

Given the above, what is the value of

\[\displaystyle \frac 1{2!}+\frac 1{3!}+\frac {a_1}{4!}+\frac {b_1}{5!}+\frac {c_1}{6!}+\frac {a_2}{7!}+\frac {b_2}{8!}+\frac {c_2}{9!}+\frac {a_3}{10!}+\frac {b_3}{11!}+\frac {c_3}{12!}+\cdots\, ?\]

**Cite as:**Telescoping Series - Sum.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/telescoping-series/