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Telescoping Series

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 \( t_{k} \). The benefit of such a series, is that it allows us to easily add up the terms, because \[ u_1 + u_2 + u_3 + \ldots + u_n = (t_1 - t_2) + (t_2 - t_3) + (t_3 - t_4 ) + \ldots + (t_n - t_{n+1} ) = t_1 - t_ {n+1}. \]

Observe that most of the \(t_k\) terms cancel out with their counterparts in other brackets, and hence 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

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

Image credit: Wikipedia

Note by Calvin Lin
3 years, 5 months ago

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You may want to update the link to Arron's Set. Here it is: https://brilliant.org/profile/arron-udsft3/sets/telescoping-series-investigation/

Also, typo: "we have \(u_k=t_k-t_{k+1}\)" Daniel Liu · 3 years, 5 months ago

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@Daniel Liu Thanks! Edited :) Calvin Lin Staff · 3 years, 5 months ago

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\((x-4),~(x -2),~x,~(x+2),~are~one~ degree,~binomiamls~in~A.P.~~with~ difference~d=2.\\
They ~are ~n=4~terms.~~ G(x)= \dfrac 1 {(x-4)(x -2)x(x+2)}.\\
If~a~series~starts~at~S=6,~and~ends~at~E=12,~~Find~~\displaystyle~\sum_{x=S}^ E G(x).\\
Apart ~from~Partial~Fraction~there~is~another~method~given~below.\\
Name~~~L~~~as~Last-term,~L=(x+2) ,~~~F~as~First-term,~F=(x-4).~~K=L-F=6.\\
P(x)=\dfrac F K*G(x),~~~~Q(x)=\dfrac L K*G(x). ~~\scriptsize~ both~ expressions ~have~ only~(n-1)=3~terms~because ~of~ cancellation.\\
\displaystyle Using~ the~ formula~\large \color{green}{\sum_{x=S}^E G(x)=\frac 1 K*\Bigg(\sum_{x=S}^{S+d-1} Q(x) - \sum_{x=E-d+1}^{E} P(x)\Bigg ).}\\
\displaystyle Substituting~given~values,~ \sum_{x=6}^{12} G(x)=\frac 1 6*\Big\{\sum_{x=6}^7 Q(x) - \sum_{x=11}^{12} P(x) \Big \}.\\
P(x)=\dfrac{x-4}{(x-4)(x-2)x(x+2)}=\dfrac 1 {(x-2)x(x+2)}.~~~Q(x)=\dfrac{x+2}{(x-4)(x-2)x(x+2)}=\dfrac 1 {(x-4)(x-2)x}.\\
\displaystyle \therefore~\frac 1 6*\sum_{x=6}^7 Q(x) =\frac 1 6*\Big \{\dfrac 1 {2*4*6}+\dfrac 1 {3*5*7} \Big \}=\dfrac{51}{18*560}...(a)\\
\displaystyle \therefore~\frac 1 6*\sum_{x=11}^{12} P(x) =\frac 1 6*\Big \{\dfrac 1 {9*11*13}+\dfrac 1 {10*12*14} \Big \}=\dfrac{\frac{989}{429}}{18*560}...(b)\\
\displaystyle \color{red}{ \sum_{x=6}^{12} \dfrac 1 {(x-4)(x -2)x(x+2)}=\dfrac{51 -\frac{989}{429}}{18*560}}=\dfrac{2089}{432432}\\
\text{Below we see how this formula is developed.}\\
G(x)=\dfrac 1 {(x-4)(x -2)x(x+ 2)}=\frac 1 {(x+2) - (x-4)}*\dfrac{ \color{orange}{(x+2)} - \color{green}{(x-4)} } { {\color{green}{(x-4)}}(x-2)x \color{orange}{(x+2)}}\\
=\frac 1 {(x+2) - (x-4)}*\Big \{{\color{orange}{\dfrac 1{ (x-4)(x-2)x} }}- {\color{green}{\dfrac 1{ (x-2)x (x+2)}}}\Big \}\\
\scriptsize \qquad \qquad \qquad \qquad ~Let ~{\color{orange}{Q(x)=\dfrac 1{ (x-4)(x-2)x} }}, ~~~{\color{green}{P(x)=\dfrac 1{ (x-2)x (x+2)}}} \\
\therefore~G(x)=\frac 1 {(x+2) - (x-4)}*\Big \{Q(x) -P(x)\Big \} \\
\implies~\displaystyle~ \sum_{x=6}^{12} G(x)=\frac 1 {(x+2) - (x-4)}*\Big \{\sum_{x=6}^{12} Q(x) - \sum_{x=6}^{12} P(x).\Big \} \\
\qquad \qquad \qquad \scriptsize \color{green}{But~~\displaystyle \sum_{n=6}^{12}Q(x)=\sum_{n=6}^7 Q(x)+ \sum_{n=6}^{10} P(x)...Proof~in~Notes ~below.}\\
= \displaystyle \frac 1 {(x+2) - (x-4)}*\Big \{\sum_{x=6}^7 Q(x) +\sum_{x=6}^{10} P(x)- \sum_{x=8}^{10} P(x) - \sum_{11}^{12} P(x).\Big \}\\
\displaystyle \frac 1 {(x+2) - (x-4)}*\Big \{\sum_{x=6}^7 Q(x) - \sum_{11}^{12} P(x).\Big \}\\ \)

TO COMPLETE SOON

\(\color{green}{Note-~~\displaystyle \sum_{n=6}^{12}Q(x) \\
\displaystyle = \sum_{n=6}^{7} Q(x) + \sum_{n=8}^{12} Q(x)\\
\displaystyle =\sum_{n=6}^{7} Q(x)+ \sum_{n=8}^{12} \dfrac 1 {(x-4)(x -2)x(x+2)}.\\
\displaystyle =\sum_{n=6}^{7} Q(x)+ \sum_{n=6}^{10} \dfrac 1 {(x -2)x(x+2)(x+4)}.\\
\displaystyle = \sum_{n=6}^{7} Q(x)+ \sum_{n=6}^{10} P(x)}. \) Niranjan Khanderia · 4 months, 3 weeks ago

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@Niranjan Khanderia You're right to point out that there is a better way in certain cases like this. Partial fractions allow us to quickly reduce to a "known problem", instead of having to "Creatively determine the telescoping terms \(t_k \)".

Note: You're missing a factor of \( \frac{1}{7} \). Calvin Lin Staff · 4 months, 3 weeks ago

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@Calvin Lin Thanks for your comment. I have added \(\frac 1 7 \).
I have changed problem so it is made more clear. Avoid two 4 in (x-4) and (x+4). Also have an upper limit so can show its working. I will add more explanation soon. Niranjan Khanderia · 4 months, 3 weeks ago

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This method of reduction is known as Partial Fraction. Often useful in integration also. Niranjan Khanderia · 3 years, 3 months ago

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@Niranjan Khanderia This is not the method of partial fractions. Please read again to clear doubts. Paramjit Singh · 3 years, 2 months ago

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@Paramjit Singh Partial-Fraction Decomposition is the full name. Niranjan Khanderia · 3 years, 2 months ago

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