Telescoping Series

A telescoping series is a series where each term uk u_k can be written as uk=tktk+1 u_k = t_{k} - t_{k+1} for some series tk t_{k} . The benefit of such a series, is that it allows us to easily add up the terms, because u1+u2+u3++un=(t1t2)+(t2t3)+(t3t4)++(tntn+1)=t1tn+1. 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 tkt_k terms cancel out with their counterparts in other brackets, and hence we are only left with t1tn+1 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 uk=1k(k+1) u_k = \frac{1}{k(k+1)} and terms tk=1k t_k = \frac{ 1}{k} , we have uk=tktk+1 u_k = t_k - t_{k+1} since

1k1k+1=k+1k(k+1)kk(k+1)=1k(k+1). \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

i=1n1i(i+1)=111n+1. \sum_{i=1}^n \frac{1}{i(i+1) } = \frac{1}{1} - \frac{ 1}{n+1}.

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

i=11i(i+1)=1. \sum_{i=1}^\infty \frac{1}{i(i+1) } = 1.

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Note by Calvin Lin
7 years, 2 months ago

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You may want to update the link to Arron's Set. Here it is:

Also, typo: "we have uk=tktk+1u_k=t_k-t_{k+1}"

Daniel Liu - 7 years, 2 months ago

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Thanks! Edited :)

Calvin Lin Staff - 7 years, 2 months ago

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This method of reduction is known as Partial Fraction. Often useful in integration also.

Niranjan Khanderia - 7 years ago

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This is not the method of partial fractions. Please read again to clear doubts.

A Brilliant Member - 6 years, 10 months ago

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Partial-Fraction Decomposition is the full name.

Niranjan Khanderia - 6 years, 10 months ago

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(x4), (x2), x, (x+2), are one degree, binomiamls in A.P.  with difference d=2.They are n=4 terms.  G(x)=1(x4)(x2)x(x+2).If a series starts at S=6, and ends at E=12,  Find   x=SEG(x).Apart from Partial Fraction there is another method given below.Name   L   as Lastterm, L=(x+2),   F as Firstterm, F=(x4).  K=LF=6.P(x)=FKG(x),    Q(x)=LKG(x).   both expressions have only (n1)=3 terms because of cancellation.Using the formula x=SEG(x)=1K(x=SS+d1Q(x)x=Ed+1EP(x)).Substituting given values, x=612G(x)=16{x=67Q(x)x=1112P(x)}.P(x)=x4(x4)(x2)x(x+2)=1(x2)x(x+2).   Q(x)=x+2(x4)(x2)x(x+2)=1(x4)(x2)x. 16x=67Q(x)=16{1246+1357}=5118560...(a) 16x=1112P(x)=16{191113+1101214}=98942918560...(b)x=6121(x4)(x2)x(x+2)=5198942918560=2089432432Below we see how this formula is developed.G(x)=1(x4)(x2)x(x+2)=1(x+2)(x4)(x+2)(x4)(x4)(x2)x(x+2)=1(x+2)(x4){1(x4)(x2)x1(x2)x(x+2)} Let Q(x)=1(x4)(x2)x,   P(x)=1(x2)x(x+2) G(x)=1(x+2)(x4){Q(x)P(x)}  x=612G(x)=1(x+2)(x4){x=612Q(x)x=612P(x).}But  n=612Q(x)=n=67Q(x)+n=610P(x)...Proof in Notes below.=1(x+2)(x4){x=67Q(x)+x=610P(x)x=810P(x)1112P(x).}1(x+2)(x4){x=67Q(x)1112P(x).}(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{#20A900}{\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{#D61F06}{ \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{#EC7300}{(x+2)} - \color{#20A900}{(x-4)} } { {\color{#20A900}{(x-4)}}(x-2)x \color{#EC7300}{(x+2)}}\\ =\frac 1 {(x+2) - (x-4)}*\Big \{{\color{#EC7300}{\dfrac 1{ (x-4)(x-2)x} }}- {\color{#20A900}{\dfrac 1{ (x-2)x (x+2)}}}\Big \}\\ \scriptsize \qquad \qquad \qquad \qquad ~Let ~{\color{#EC7300}{Q(x)=\dfrac 1{ (x-4)(x-2)x} }}, ~~~{\color{#20A900}{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{#20A900}{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 \}\\


Note  n=612Q(x)=n=67Q(x)+n=812Q(x)=n=67Q(x)+n=8121(x4)(x2)x(x+2).=n=67Q(x)+n=6101(x2)x(x+2)(x+4).=n=67Q(x)+n=610P(x).\color{#20A900}{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 years, 1 month ago

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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 tkt_k ".

Note: You're missing a factor of 17 \frac{1}{7} .

Calvin Lin Staff - 4 years, 1 month ago

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Thanks for your comment. I have added 17\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 years, 1 month ago

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