Evaluating the infinite series $\sum\limits_{n=1}^\infty(\sin\frac1{2n}-\sin\frac1{2n+1})$

Solution 1:

$$\sum_{n=1}^\infty \sin\left({1\over 2n}\right)-\sin\left({1\over 2n+1}\right)=-\sum_{n=1}^\infty\int_{1\over 2n+1}^{1\over 2n}\cos x\, dx$$

These are integrals of a bounded function on the sets

$$\left[{1\over 2n+1},{1\over 2n}\right]$$

The measure of these sets is

$${1\over 2n}-{1\over 2n+1}={1\over 2n(2n+1)}$$

Hence the sum is absolutely convergent by Jensen's inequality since

$$\left|\sum_{n=1}^\infty\int_{1\over 2n}^{1\over 2n+1}\cos x\right|\le\sum_{n=1}^\infty\int_{1\over 2n}^{1\over 2n+1}|\cos x|\le\sum_{n=1}^\infty\int_{1\over 2n+1}^{1\over 2n}1\;dx=\sum_{n=1}^\infty {1\over 2n(2n+1)}.$$

Solution 2:

I can't give precise value for the sum of the series, but note that the Mean Value Theorem tells us that it may be written in the form $\sum_{n=1}^{\infty} \frac{ \cos(\theta_{2n})}{2n(2n+1)},$ where each $\theta_{2n} \in (\frac{1}{2n+1},\frac{1}{2n})$ which seems to mean that it differs from $ 1 - \log2 = \sum_{n=1}^{\infty}\frac{1}{2n(2n+1)}$ by something close to $\sum_{n=1}^{\infty}\frac{1}{32n^{4}} = \frac{\pi^{4}}{2880}$, (and is smaller than that sum).