I've been bored and came across in my book a pretty straightforward series problem, namely to determine the convergence of
$$
\sum_{n = 1}^{\infty}
\left[\sin\left(1 \over 2n\right) – \sin\left(1 \over 2n + 1\right)\right]
$$
Doing so was trivial by rewriting it as an alternating series involving the term $(-1)^k\sin\frac1k$.
Naturally, though, I was curious as to whether this series can be reduced to a simpler closed form in terms of more fundamental constants. Unfortunately I do not immediately know of any techniques of use here or even whether it permits such a 'nice' form. Do any of you?
I do know from playing with the Euler-Maclaurin sums the value should be something near $0.290674$. As $n\to\infty$ I know the sequence terms behave increasingly like those of the alternating harmonic series (as $\sin x\sim x$ for $|x|\ll1$), which helps explain why it appears relatively near $1-\log2$. I have also found that the difference between it and the alternating harmonic series starting with $1/2$ is near $0.016179$.
I should note that I am a high school student with an amateur interest in recreational math. My knowledge extends only as far as elementary calculus of multiple variables and first-year ordinary and partial differential equations. It may very well be that an obvious approach exists that I've completely missed and so I feel obligated to apologize in advance.
Best Answer
$$\sum_{n=1}^\infty \sin\left({1\over 2n}\right)-\sin\left({1\over 2n+1}\right)=-\sum_{n=1}^\infty\int_{1\over 2n+1}^{1\over 2n}\cos x\, dx$$
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
$$\left|\sum_{n=1}^\infty\int_{1\over 2n}^{1\over 2n+1}\cos x\right|\le\sum_{n=1}^\infty\int_{1\over 2n}^{1\over 2n+1}|\cos x|\le\sum_{n=1}^\infty\int_{1\over 2n+1}^{1\over 2n}1\;dx=\sum_{n=1}^\infty {1\over 2n(2n+1)}.$$