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My friend was practicing for his entrance examination and one of the problems on older exams was this one. It asks if it converges and to explain the reasoning behind that answer. The series is:

$$\sum_{n=1}^\infty \sin^2\left(\pi(n + \tfrac{1}{n})\right)$$

We had a few ideas, but none of them seemed to be working. Hoping for some help!

Fly by Night
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hadsed
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3 Answers3

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Using $\lim_{x \to 0} \frac{\sin x}{x} = 1$ we can write $$\sin^2 \left(\pi \left( n+\frac{1}{n} \right) \right) = \sin^2 \frac{\pi}{n} \sim \frac{\pi^2}{n^2}, \quad n \to \infty.$$ Here $f \sim g$ means $\frac{f}{g} \to 1$ (and thus is bounded from both sides). Since the series $\sum \frac{1}{n^2}$ converges (actually $= \frac{\pi^2}{6}$) by the comparison test your series also converges.

njguliyev
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    This answer would be much better if you defined $\sim$ and if you justified the relation to $\tfrac{\pi^2}{n^2}$. After all, if the OP already knows what $\sim$ means and can already justify the relation to $\tfrac{\pi^2}{n^2}$ then s/he would have no need of posting this question. – Fly by Night Aug 18 '13 at 20:41
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    @Alex, $f \sim g$ means $\frac{f}{g} \to 1$. – njguliyev Aug 18 '13 at 21:39
  • @njguliyev You could edit your reply to include this information. I don't think that breaks any rules :o) – Fly by Night Aug 18 '13 at 22:02
  • @FlybyNight, done. – njguliyev Aug 18 '13 at 22:14
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    @Alex Sorry? An upper bound by the general term of an absolutely convergent series is definitely enough. – Did Aug 18 '13 at 22:33
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We have $$\sin^2\left(\pi\left(n+\frac 1 n\right)\right)=\sin^2\left(\frac{\pi}{n}\right)=\frac{\pi^2}{n^2}+o\left(\frac{1}{n^2}\right)$$ so we conclude that the given series is convergent by comparison with the Riemann series.

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    This answer would be much better if you justified the final equality. After all, if the OP understands how you derived it, then why would s/he have asked the question? – Fly by Night Aug 18 '13 at 20:43
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    ...or the other way around, @FlybyNight: whoever knows about Taylor/power series knows how the last equality is derived, or else he can work it out by himself, and whoever doesn't know then the answer's not for him. I think this is a nice answer.+1 – DonAntonio Aug 18 '13 at 21:37
  • @DonAntonio That's the problem. If the reader has a good understanding of power series - so good that they can recongise their use without them being mentioned - then the answer is of use. But then again, if someone is so familiar with power series that they can spot them being used without them being mention, don't you think they would have tried that by themselves already? It might be that my philosophy about this site is all wrong. I thought we were in the business of giving thoughtful, helpful well-crafted, accessible replies and not grandstanding our own mathematical learning. – Fly by Night Aug 18 '13 at 21:47
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    Imo you think too much all this, @FlybyNight . Power series are not that advanced a subject wrt series as to be impossible to think that people can have problems with them even after knowing both. I don't know if this is the case, but even if the OP doesn't really know power series this nice answer can help others. And no: imo, we're not in that business you say. Anyway I am not. I'm here because I'm a mathematics freak and I love to deal with mathematics, and I see my role here as someone giving hints, indications, etc. to others so that they can develop by themselves their own answers – DonAntonio Aug 18 '13 at 21:52
  • @DonAntonio I sympathise with your last few sentences. But either way, simply saying "We can use power series to show that..." would not have been very difficult and would have made the reply much better. That was my original point. Even if you only supply hits, or whatever, you want them to be meaningful and purposeful. – Fly by Night Aug 18 '13 at 22:01
  • Well, in remaking the use of power series I think I could agree with you, @FlybyNight . – DonAntonio Aug 18 '13 at 22:05
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  • For every $x$ in $[0,\pi]$, $0\leqslant\sin(x)\leqslant x$ hence $\sin^2(x)\leqslant x^2$.

  • For every $n\geqslant1$, $\sin^2(\pi(n+1/n))=\sin^2(\pi/n)$ and $\pi/n$ is in $[0,\pi]$ hence $\sin^2(\pi/n)\leqslant\pi^2/n^2$.

  • Finally, the (Riemann) series $\sum\limits_{n\geqslant1}1/n^2$ converges hence, by comparison of series with nonnegative terms, the series $\sum\limits_{n\geqslant1}\sin^2(\pi(n+1/n))$ converges (and its sum is at most $\pi^4/6$).

Did
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