# Telescoping Series/Example 1

## Theorem

Let $\left \langle {b_n} \right \rangle$ be a sequence in $\R$.

Let $\left \langle {a_n} \right \rangle$ be a sequence whose terms are defined as:

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

Then:

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

If $\left \langle {b_n} \right \rangle$ converges to zero, then:

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

## Proof

 $\displaystyle \displaystyle \sum_{k \mathop = 1}^n a_k$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \left({b_k - b_{k + 1} }\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 1}^n b_{k + 1}$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 2}^{n + 1} b_k$ Translation of Index Variable of Summation $\displaystyle$ $=$ $\displaystyle b_1 + \sum_{k \mathop = 2}^n b_k - \sum_{k \mathop = 2}^n b_k - b_{n + 1}$ $\displaystyle$ $=$ $\displaystyle b_1 - b_{n + 1}$

If $\left \langle {b_k} \right \rangle$ converges to zero, then $b_{n + 1} \to 0$ as $n \to \infty$.

Thus:

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

So:

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

$\blacksquare$

## Linguistic Note

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