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

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

## Comments

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TopNewestYou 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 · 2 years, 9 months ago

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– Calvin Lin Staff · 2 years, 9 months ago

Thanks! Edited :)Log in to reply

This method of reduction is known as Partial Fraction. Often useful in integration also. – Niranjan Khanderia · 2 years, 7 months ago

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– Paramjit Singh · 2 years, 6 months ago

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

– Niranjan Khanderia · 2 years, 6 months ago

Partial-Fraction Decomposition is the full name.Log in to reply