# infinite sums of trigonometric functions

Find the sum of the series: $$\sum_{n = 1}^\infty \left( \sin \left(\frac{1}{n}\right) - \sin\left(\frac{1}{n+1} \right) \right).$$

By the convergence test the limit of this function is $0$ but I'm not sure how to prove whether or not this function converges or diverges.

-

## 1 Answer

Write the sum as $\sin(\frac{1}{1}) - \sin(\frac{1}{2}) + \sin(\frac{1}{2}) - \sin(\frac{1}{3}) + \sin(\frac{1}{3}) - \sin(\frac{1}{4}) + \cdots$. All terms but the first cancel and we are left with $\sin(1)$. You have already established that the limit of the terms is $0$, so the limit of the sum is $\sin(1)$.

-
In other words it's a telescoping sum. one may google that term. –  Michael Hardy Jul 17 '12 at 3:02