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

I've been bored and playing with infinite series and came across in my book the following 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 a 'simpler' summation involving the general term $(-1)^k\sin\frac1k$ which is trivial to prove convergence for as a simple alternating series. Naturally, though, I was curious to determine if there was a means of actually evaluating the series. Unfortunately, I do not yet know of a technique to determine what this converges to in terms of more fundamental constants (does it converge to a 'nice' value?)

I do know from playing with the Euler-Maclaurin summation formula and playing with partial sums the value should be something near $0.290674$. As $n\to\infty$ I know the sequence terms behave more and more like those of the alternating harmonic series (given $\sin x\sim x$ for $|x|\ll1$), which explains why it gives a value relatively near $1-\log2$ but that's about as far as I know. I have also found that the difference between it and the alternating harmonic series starting with $1/2$ is near $0.016179$.

Does anyone know of a way to evaluate the above series? Does it permit a nice closed-form expression for its value? Thank you ahead of time.

I should note that I am a math novice and a high school student. My math knowledge only encompasses only that which fits elementary calculus of single and multiple variables and first year ordinary and partial differential equations. It may be that a very simple approach exists that I totally missed and so I feel obligated to apologize ahead of time.

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Using the trigonometric identity for $\sin(a)-\sin(b)$ might be helpful . – Zaid Alyafeai Aug 22 '13 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. – oldrinb Aug 22 '13 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... – poirot Aug 22 '13 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. – Daniel Fischer Aug 22 '13 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. – Daniel Fischer Aug 23 '13 at 11:29

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).

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

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

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

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Let us recall the fact that If $a_k$ is a sequence of real numbers which converges to $0$, then $\sum_{k=1}^{\infty} \left({a_k - a_{k+1}}\right) = a_1$

Proof.

Utilizing the sequence of partial sums of the series $\sum_{k=1}^{\infty} \left({a_k - a_{k+1}}\right)$ we have $S_n =\sum_{k=1}^{n} \left({a_k - a_{k+1}}\right)$

\begin{align*} S_n &=\sum_{k=1}^{n} \left({a_k - a_{k+1}}\right) \\ &= \left({a_1 - a_{2}}\right) + \left({a_2 - a_{3}}\right) + ...+\left({a_{n-1} - a_{n}}\right)+\left({a_n - a_{n+1}}\right) \\ &= a_1 - a_{n+1} \end{align*} and since $a_n \to 0$ then $\mathop {\lim }\limits_{n \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } (a_1 - a_{n+1}) = a_1$.

Applying this fact for the sequence $a_k:=\sin (\frac{1}{2k})$, we get that
\begin{align*} \sum\limits_{n = 1}^\infty {\left( {\sin \frac{1}{{2n}} - \sin \frac{1}{{2n + 1}}} \right)} = \sin \left( {\frac{1}{2}} \right) \end{align*}

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But here we have $\sum (a_{2k} - a_{2k+1})$, or, if you wish, $\sum (a_k - a_{k+\frac{1}{2}})$, it doesn't telescope. – Daniel Fischer Apr 28 '14 at 11:25
Ah yes. You are right. But how come $k+\frac{1}{2}$ in $a_{k+\frac{1}{2}}$ – mwomath May 6 '14 at 11:18
@user45105 $a_k=\sin(1/(2k))\implies a_{k+1/2}=\sin(1/(2(k+1/2)))=\sin(1/(2k+1))$ – oldrinb May 7 '14 at 22:19
If we substitute $m=2n$, then \begin{align} \sum\limits_{n = 1}^\infty {\left( {\sin \frac{1}{{2n}} - \sin \frac{1}{{2n + 1}}} \right)} &= \sum\limits_{m = 2}^\infty {\left( {\sin \frac{1}{m} - \sin \frac{1}{{m + 1}}} \right)} \\ &= \sum\limits_{m = 1}^\infty {\left( {\sin \frac{1}{m} - \sin \frac{1}{{m + 1}}} \right)} - \left( {\sin 1 - \sin \frac{1}{2}} \right) \\ &= \sin 1 - \left( {\sin 1 - \sin \frac{1}{2}} \right)\\ &= \sin \frac{1}{2} \end{align} where as above $\sum\limits_{m = 1}^\infty {\left( {\sin \frac{1}{m} - \sin \frac{1}{{m + 1}}} \right)} = \sin \left( 1 \right)$ – mwomath Jun 3 '14 at 10:59