# Telescoping Series/Example 1

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## Theorem

Let $\sequence {b_n}$ be a sequence in $\R$.

Let $\sequence {a_n}$ be a sequence whose terms are defined as:

- $a_k = b_k - b_{k + 1}$

Then:

- $\ds \sum_{k \mathop = 1}^n a_k = b_1 - b_{n + 1}$

If $\sequence {b_n}$ converges to zero, then:

- $\ds \sum_{k \mathop = 1}^\infty a_k = b_1$

## Proof

\(\ds \ds \sum_{k \mathop = 1}^n a_k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {b_k - b_{k + 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 1}^n b_{k + 1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 2}^{n + 1} b_k\) | Translation of Index Variable of Summation | |||||||||||

\(\ds \) | \(=\) | \(\ds b_1 + \sum_{k \mathop = 2}^n b_k - \sum_{k \mathop = 2}^n b_k - b_{n + 1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds b_1 - b_{n + 1}\) |

If $\sequence {b_k}$ converges to zero, then $b_{n + 1} \to 0$ as $n \to \infty$.

Thus:

- $\ds \lim_{n \mathop \to \infty} s_n = b_1 - 0 = b_1$

So:

- $\ds \sum_{k \mathop = 1}^\infty a_k = b_1$

$\blacksquare$

## Linguistic Note

The term **telescoping series** arises from the obvious physical analogy with the folding up of a telescope.