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

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.

• Using the trigonometric identity for $\sin(a)-\sin(b)$ might be helpful . Aug 22, 2013 at 20:55
• I tried that @ZaidAlyafeai but unfortunately I could not determine anything neat. I will look into it again however. I must leave for class though as it begins in a few minutes. Thank you for your response. Aug 22, 2013 at 20:56
• Have you tried considering the imaginary part of $e^{i\theta} = \cos\theta + i\sin\theta$ ? I don't know if this will help, but might be worth a try...
– gone
Aug 22, 2013 at 21:00
• If you just expand the sums, you get (if I haven't miscalculated) $$1 - \log 2 + \sum_{k=1}^\infty \frac{(-1)^k\bigl(1-(1-2^{-2k})\zeta(2k+1)\bigr)}{(2k+1)!} = \sin 1 - \log 2 +\sum_{k=1}^\infty \frac{(-1)^{k+1}(1-2^{-2k})\zeta(2k+1)}{(2k+1)!},$$ which isn't a nice closed-form expression, but at least involves everybody's favourite, the Riemann $\zeta$-function. Aug 22, 2013 at 21:10
• For me, both formulae produce $0.2906741\ldots$. Although I haven't taken any measures to minimise floating point errors, those should not affect the given significant digits. Aug 23, 2013 at 11:29

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

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

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

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).