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I have to test for convergence of the series:

  • $\displaystyle \sum\limits_{n=1}^{\infty} \sin\Bigl(\frac{\pi}{n}\Bigr)$

What i did was

\begin{align*} \sin\Bigl(\frac{\pi}{n}\Bigr)+ \sin\Bigl(\frac{\pi}{n+1}\Bigr) + \cdots & < \pi \biggl( \frac{1}{n+1} + \frac{1}{n+2} + \cdots \biggr) \\ &= \pi \biggl( \sum\limits_{r=1}^{\infty} \frac{1}{n+r}\biggr) =\int\limits_{0}^{1} \frac{1}{1+x} \ dx \\ &= \pi\log{2} \end{align*}

I think this proves the convergence of the series.

  • I am interesting in knowing some more methods which can be used to prove the convergence so that i can apply them.

ADDED: Note that $\lim_{n \to \infty} \sum\limits_{r=1}^{n} \frac{1}{n} \cdot f\Bigl(\frac{r}{n}\Bigr) = \int\limits_{0}^{1} f(x) \ \textrm{dx}$

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    Nobody has mentioned the standard calculus theorem here; usually called the "limit comparison test" (at least e.g. in Stewart's calculus textbook). More generally as a consequence of this test, if $a_n$ is any nonnegative sequence with limit $0$, $\sum \sin(a_n)$ converges if and only if $\sum a_n$ converges (for the reason that $\frac{\sin(t)}{t}$ goes to $1$ as $t \to 0$). Perfectly rigorous and no harder to prove than the standard "comparison test" for series of nonnegative terms.2010-12-11

2 Answers 2

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You observed correctly that for small angles $\frac{\pi}{n}$, sin$\left(\frac{\pi}{n}\right)$ is very close to $\frac{\pi}{n}$. As for convergence of $\pi\sum_{n=1}^\infty\frac{1}{n}$...

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    @Willie Yes, this was not an attempt at a rigorous answer. It is actually quite easy to explicitly bound the terms below, so no need for error analysis. But I will leave that to Chandru.2010-12-10
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The way of solving the problem is wrong. We know that $\sum \frac1n$ is divergent series and if we apply the comparison test limit form in $\sum \frac1n$ and $\sum \sin\Big(\frac1n \Big)$ we show that both the series diverge or converge together. then the given series is divergent.