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I was trying to determine whether or not the following sum, $$\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)$$, would converge or diverge. I thought that this function would diverge, as it is comparable to the harmonic series, which is $$\sum_{n=1}^{\infty}{\frac{1}{n}}$$, but I remembered that the ranges from 0 to 1. So, it can be viewed as having $(-1)^n$, which means this sum would converge as $$\lim_{n\to\infty}{\frac{\sin(n)}{n}} = 0$$

So, what value would the sum converge to?

Jaden Lee
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  • It is comparable and it can be viewed is pretty loose language for math :) The answer is $1/2 \cdot (\pi-1)$ – peter.petrov Apr 17 '20 at 00:16
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    @rtybase yeah. Apparently, I did not look hard enough to find the solution to my question. Thanks! – Jaden Lee Apr 17 '20 at 00:24
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    The partial sums $\sum_{n=1}^p \sin(n)$ is bounded and $\frac1n$ monotonic decreases to $0$. You can apply Dirichlet's test to conclude the series conditonally converges. When a series $\sum_{k=0}^\infty a_k$ conditionally converges, you can apply Abel's theorem to evaluate it $$\sum_{k=0}^\infty a_k = \lim_{t\to 1^-} \sum_{k=0}^\infty a_k t^k$$ For this series, $\sin(n) = \Im e^{in}$ and $-\log(1-z) = \sum_{n=1}^\infty \frac{z^n}{n}$ for $|z| < 1$, ... – achille hui Apr 17 '20 at 00:43

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