Telescoping Series/Example 2

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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:

$\displaystyle \sum_{k \mathop = m}^n a_k = b_n - b_{m - 1}$


Proof

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

$\blacksquare$


Linguistic Note

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


Sources