Telescoping series

In mathematics, a telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. For example, let's try to find value of the series $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + ... + \frac{1}{(n-1) \cdot n}$ . We can see that $\frac{1}{n-1} - \frac{1}{n} = \frac{1}{(n-1) \cdot n}$ . Thus, $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + ... + \frac{1}{(n-1) \cdot n}$ = $\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{n-1} - \frac{1}{n}$ . Then, we can see that all of the terms except $\frac{1}{1}$ and $\frac{1}{n}$ . So the answer is $1 - \frac{1}{n} = \frac{n-1}{n}$ . We can see that in the process, we manipulated a large series so the many terms cancelled out with each other, leaving only a few terms that we could easily calculate with. This is usually how most telescoping series work.

Problems

Intermediate

Find the value of $\sum_{k=2}^{\infty} \left( \zeta(k) - 1 \right),$ where $\zeta$ is the Riemann zeta function $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}.$

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Telescoping series

Problems

Intermediate