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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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17  
How did you manage to sum the harmonic series? –  Willie Wong Dec 10 '10 at 15:20
6  
Anyway, the series you have diverges. –  Andres Caicedo Dec 10 '10 at 15:54
1  
@Andres Caicedo: Thanks for pointing out! –  anonymous Dec 10 '10 at 15:55
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$$\lim_{n\to\infty} \sum_{r = 1}^n\frac{1}{n} \frac{1}{1 + \frac{r}{n}} \neq \sum_{r = 1}^\infty \frac{1}{n+r}$$ The left hand side is independent of $n$, the right hand side isn't. There's a piece $\lim_{n\to\infty}\sum_{r = n+1}^\infty \frac{1}{n+r}$ that you are missing when you "applied the formula". If that term were convergent, then it would go away when you take the limit. Unfortunately, the series inside is divergent... –  Willie Wong Dec 10 '10 at 16:17
5  
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. –  anon Dec 11 '10 at 7:45
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2 Answers

up vote 15 down vote accepted

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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4  
Ah, you beat me to my comment above. Also to be slightly more rigorous, one should note that the error term from approximating $\sin(\pi/n)$ by $\pi/n$ for large $n$ is $O(n^{-2})$, so the error term gives finite contribution to the sum. –  Willie Wong Dec 10 '10 at 15:22
3  
@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. –  Alex B. Dec 10 '10 at 15:31
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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.

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