Question

Show that the series \(\sum_{n=1}^{\infty} a_{n}\) can be written in the telescoping form \(\sum_{n=1}^{\infty}\left[\left(c-S_{n-1}\right)-\left(c-S_{n}\right)\right]\) where \(S_{0}=0\) and \(S_{n}\) is the \(n\) th partial sum.

Short answer

The series can be written in telescoping form because, in the process of addition (series), each term \(S_{n-1}\) cancels the previous term \(-S_{n-1}\). As a result, we're left with the original series, demonstrating that it can be written in such a manner.

Step-by-step solution

Recall the Concept of Partial Sum

It is critical to initially remember the idea of a partial sum. In a given series, the ith partial sum \(S_{i}\) is the sum of the first i terms. In this way, by this definition, \(S_{0}\) ought to be 0 as it is the 'sum' of zero terms and each successive \(S_{i}\) is identified by the sum of \(S_{i-1}\) (the (i-1)th partial sum) and \(a_{i}\) (the ith term of the series). So we have: \(S_{i} = S_{i-1} + a_{i}\).

Rewrite the Series in the telescoping form

First, let's draw attention to the telescoping series. Here we are dealing with \[\sum_{n=1}^{\infty}\left[\left(c-S_{n-1}\right)-\left(c-S_{n}\right)\right]\] If we distribute the subtraction sign in the brackets, this becomes \[\sum_{n=1}^{\infty}(c - S_{n-1} - c + S_{n})\] The two 'c' values cancel out and now we are left with \[\sum_{n=1}^{\infty}(S_{n} - S_{n-1})\]

Relate the telescoping form to the original series

It’s visible that the terms now follow a pattern wherein every term \(S_{n-1}\) cancels the previous term \(-S_{n-1}\) for n > 1. So, when we sum everything, all these terms cancel out and we are left with the terms from the sequence that could not cancel out, which are \(S_{n}\) terms and the first \(S_{n-1}\) term, which equals 0. So what’s left is \(\sum_{n=1}^{\infty} a_{n}\). Hence, the proof.