$$\sum_{n=1}^\infty \sin^2(\frac{\pi}{n}) $$

First, I tried Divergence test, but the limit is $0$, so that's inconclusive. I don't think I can do integral test, since it's not a clean integration.

Did I do this right ? If it's wrong, can you steer me in the right direction? I tried subbing in $1-\cos(x)$ for the $\sin(x)$. This would get:

$$\sum_{n=1}^\infty \sin^2(\frac{\pi}{n})$$ $$= \sum_{n=1}^\infty 1-\cos^2(\frac{\pi}{n})$$ $$ = \sum_{n=1}^\infty 1 - \sum_{n=1}^\infty \sin^2(\frac{\pi}{n})$$ But that first sigma is divergent to infinity, so does the original series also diverge to infinity?

The tests I know are: Geometric, p-series, Divergence (nth term) test, Integral test, Direct Comparison test, Alternating Series Test, Absolute convergence, and Ratio test.

  • $\begingroup$ You third from last line is illogical. Just the fact that a infinite sum can be split in to two diverging sums does not mean that the original infinite sum is divergent. $\endgroup$
    – S.C.B.
    Mar 29 '16 at 16:34

Using $\sin(x)\le x$ for all $x\ge0$ so by comparison with the convergent series $\sum\frac1{n^2}$ we conclude the convergence of the given series.


Use the limit comparison test:

$$\frac{\sin^2\frac\pi n}{\frac{\pi^2}{n^2}}=\left(\frac{\sin\frac\pi n}{\frac\pi n}\right)^2\xrightarrow[n\to\infty]{}1^2=1$$

and thus your series converges since $\;\sum\frac{\pi^2}{n^2}\;$ does.

  • $\begingroup$ That was an easy one. $\endgroup$ Mar 29 '16 at 16:39
  • $\begingroup$ How did you get $1^2$ ? I get indeterminate form of $\frac{0}{0}$ Did you use L'hopitals ? $\endgroup$
    – JackOfAll
    Mar 29 '16 at 16:59
  • $\begingroup$ @Jack: you know that $\;\frac{\sin(f(x))}{f(x)}\xrightarrow[x\to x_0]{}1\;$ if $\;\lim\limits_{x\to x_o}f(x)=0\;$ , right? This is a basic limit $\endgroup$
    – DonAntonio
    Mar 29 '16 at 17:03
  • $\begingroup$ Oh, right, $\frac{sin(x)}{x}$ I didn't recognize it in that "disguised form" ! But, the limit is going to infinity, not 0.,.... $\endgroup$
    – JackOfAll
    Mar 29 '16 at 17:04
  • $\begingroup$ No. The argument of sine is going to zero: $$\frac\pi n\xrightarrow[n\to\infty]{}0$$ $\endgroup$
    – DonAntonio
    Mar 29 '16 at 17:10

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