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Question: Find the sum of the series: $$\lim_{n \to \infty}\frac{\sin1}{1}+\frac{\sin2}{2}+\frac{\sin3}{3}+...+\frac{\sin n}{n}$$

I have no clue how to find this. Obviously I can see the sum will be convergent as the denominator gets increasingly bigger while the numerator is bound between $1$ and $-1$.

Gummy bears
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    Not so obvious, in my opinion. For instance $\sum_{n=1}^\infty\frac1n$ is divergent. You might need a sharper argument. –  May 17 '16 at 05:27
  • @G.Sassatelli Aha.... Well I guess intuition can be wrong sometimes. – Gummy bears May 17 '16 at 05:31
  • Could you let us know where you have stumbled upon this problem? Knowing this might help us to find a solution which avoids tools too complicated for you. – Wojowu May 17 '16 at 07:19
  • @Wojowu Well... it's this problem for college entrances. So it should be solved using 12th grade knowledge. – Gummy bears May 17 '16 at 07:45

4 Answers4

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Alternative Hint: consider the Fourier series of the $2\pi$-periodic function that equals $\frac{\pi-x}{2}$ over $(0,2\pi)$.

Yet another way is to use the limit case of the Euler-Maclaurin summation formula: since $\text{sinc}(x)$ is a very well-behaved even analytic function, we simply have: $$ \sum_{n\geq 1}\frac{\sin n}{n}=\int_{0}^{+\infty}\frac{\sin x}{x}\,dx-\frac{1}{2}.$$

Jack D'Aurizio
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  • 'Very well-behaved even analytic function'? What does it mean for a function to be 'well-behaved'? xD And how would i go about solving that integral? It doesn't seem simple... – Gummy bears May 17 '16 at 05:56
  • You may find many topics here on MSE showing that $\int_{0}^{+\infty}\frac{\sin x}{x},dx=\frac{\pi}{2}$, it is a very well-known integral. – Jack D'Aurizio May 17 '16 at 06:00
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Here's a step-by-step breakdown using the fact that $\sin x = \text{Im}(e^{ix})$

$$\lim_{n \to \infty} \sum_{k=1}^n \frac{\sin k}{k}$$ $$=\lim_{n \to \infty} \sum_{k=1}^n \frac{\text{Im}(e^{ik})}{k}$$ $$=\text{Im}\left(\sum_{k=1}^\infty \frac{e^{ik}}{k}\right) =\text{Im}\left(\sum_{k=1}^\infty \frac{(e^i)^k}{k}\right)$$ Now recall that the Taylor Series for $-\log(1-x)$ is $\sum_{k=1}^\infty \frac{x^k}{k}$
$$=\text{Im}(-\log(1-e^i))$$ $$=\text{Im}(\text{Arg}(1-e^i))$$ We now express $1-e^i$ as $1-\cos(1)-i\sin(1)$. Noting that $1-\cos(x) > 0 \;(\forall x \in \mathbb R)$ we can simply let $\text{Arg}(1-e^i) = \arctan\left(\frac{\sin(1)}{1-\cos(1)}\right) = \arctan\left(\cot(\frac{1}{2})\right)$ $$=\color{red}{\frac{1}{2}(\pi - 1)}$$

Brevan Ellefsen
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  • Just asking: is the last series convergent at $e^i$? – Chip May 17 '16 at 05:55
  • @Chip a good question. The simplest answer is that the absolute value of $e^i = 1$, which is within the interval of convergence of the series, namely $|x| \leq 1$. Now, I've mostly self studied mathematics so I am sure there are a few caveats here (there always are with complex numbers) but I am unsure of them... nevertheless, I know the formula works here. – Brevan Ellefsen May 17 '16 at 06:02
  • @Chip I found the following quote from a related post giving more justification: "Let's first observe that $\sum_{k=1}^\infty u^k/k=-\ln(1-u)$. If we're concerned about the convergence radius, we can always replace $u$ with $ue^{-\epsilon}$ and let $\epsilon\rightarrow0$. The branch of $\ln$ we're using is the one defined on $\mathbb{C}\setminus(-\infty,0]$: i.e. $\ln(re^{i\theta})=\ln r+i\theta$ where $r>0$ and $\theta\in(-\pi,\pi)$" – Brevan Ellefsen May 17 '16 at 06:07
  • many thanks! It helped. – Chip May 17 '16 at 08:29
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Hint: $\sin n$ is the imaginary part of $e^{i n}$.

Semiclassical
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It's a simple application of Abel-Plana formula

\begin{align} \sum_{n = 1}^{\infty}{\sin\left(n\right) \over n} & = -1 + \sum_{n = 0}^{\infty}{\sin\left(n\right) \over n} \\[3mm] &= -1 + \left[\int_{0}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x + {1 \over 2}\,\lim_{x \to 0}{\sin\left(x\right) \over x} + {\rm i}\int_{0}^{\infty} {\sin\left({\rm i}n\right)/\left({\rm i}n\right) - \sin\left(-{\rm i}n\right)/\left(-{\rm i}n\right)\over {\rm e}^{2\pi t} - 1}\,{\rm d}t\right] \\[3mm] &= -1 + {\pi \over 2} + {1 \over 2} + 0 = \color{#f00}{{1 \over 2}\left(\pi - 1\right)} \end{align}

Felix Marin
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