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Convergence of $$\sum_{n=1}^{\infty} \log~ ( n ~\sin \dfrac {1 }{ n })$$

Attempt:

Initial Check : $\lim_{n \rightarrow \infty } \log~ ( n ~\sin \dfrac {1 }{ n }) = 0$

$\log~ ( n ~\sin \dfrac {1 }{ n }) < n ~\sin \dfrac {1 }{ n }$

$\implies \sum_{n=1}^{\infty} \log~ ( n ~\sin \dfrac {1 }{ n }) < \sum_{n=1}^{\infty} n ~\sin \dfrac {1 }{ n }$

But, $\sum_{n=1}^{\infty} n ~\sin \dfrac {1 }{ n }$ is itself a divergent sequence. Hence, I don't think the above step is of any particular use.

Could anyone give me a direction on how to move ahead.

EDIT: I did think of trying to use the power series of $\sin$. Here's what I attempted :

$\sin \dfrac {1}{n} = \dfrac {1}{n}-\dfrac {1}{n^3.3!}+\dfrac {1}{n^5.5!}-\cdots$

$\implies n \sin \dfrac {1}{n} = 1- \dfrac {1}{n^2.3!}+\dfrac {1}{n^4.5!}-\cdots$

I couldn't proceed further due to the $\log$.

Thank you for your help.

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  • $\begingroup$ dfrac and displaystyle are not appropriate for titles you can find more about this on meta. $\endgroup$ – dustin Jan 25 '15 at 21:45
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    $\begingroup$ @Wanderer $\lim_{n\to \infty} n\sin \frac{1}{n} = 1$, so $\lim_{n\to \infty} \log(n\sin \frac{1}{n}) = 0$. $\endgroup$ – kobe Jan 25 '15 at 21:47
  • $\begingroup$ @kobe Could you please explain further. I mean, I have already used this result above. $\endgroup$ – MathMan Jan 25 '15 at 21:48
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    $\begingroup$ You used the inequality $\log x<x$, which is way too weak to be of use. When $x\approx1$, $\log x\approx0$ after all. Question: Do you know the Taylor series for $\sin x$? $\endgroup$ – Harald Hanche-Olsen Jan 25 '15 at 21:49
  • $\begingroup$ @Wanderer We have $$\lim_{n\to \infty} n\sin \frac{1}{n} = \lim_{n\to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} = \lim_{x\to 0} \frac{\sin x}{x} = 1$$ Hence $$\lim_{n\to \infty}\log(n\sin\frac{1}{n}) = \log(\lim_{n\to \infty} n\sin \frac{1}{n}) = \log(1) = 0$$ $\endgroup$ – kobe Jan 25 '15 at 21:50
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A slightly different way than the two current answers, dealing mostly with intuition:

$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots$$ or alternately, $$\sin\left(\frac 1x \right) = \frac{1}{x} - \frac{1}{3!x^3} + \frac{1}{5!x^5} + \ldots$$ Multiplying by $x$ gives that $$x\sin\left( \frac 1x \right) = 1 - \frac{1}{3!x^2} + \ldots \approx 1 - \frac{1}{6x^2},$$ and where the approximation is extremely good for high $x$ (error $\ll \frac{1}{x^4}$).

You are now interested in how $\log \left( 1 - \frac{1}{6x^2}\right)$ behaves. Fortunately, we know that $$ \log (1 - x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots$$ and thus $$ \log \left( 1 - \frac{1}{6x^2} \right) = \frac{1}{6x^2} + \ldots \approx \frac{1}{6x^2},$$ where the error is moderately good for large $x$ (no worse than $\frac{1}{x^2}$ for sufficiently large $x$).

So at long last, you are wondering about the sum $$ \sum_{x \geq 1} \frac{1}{6x^2} = \frac{\pi^2}{36},$$ though the particular value doesn't matter, but only the fact that it converges.

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    $\begingroup$ As a side note, if you are sufficiently versed in these expansions, then you can do this heuristic quickly and in your head. $\endgroup$ – davidlowryduda Jan 25 '15 at 22:06
  • $\begingroup$ Minor nit: Where you write (error $\ll\frac1{x^4}$), the $\ll$ is a bit too strong. $\endgroup$ – Harald Hanche-Olsen Jan 26 '15 at 10:57
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It is interesting to notice that, since: $$ \sin z = z\prod_{n\geq 1}\left(1-\frac{z^2}{n^2 \pi^2}\right)\tag{1}$$ we have: $$ \log\left(m\sin\frac{1}{m}\right)=\sum_{n\geq 1}\log\left(1-\frac{1}{\pi^2 n^2 m^2}\right)=-\sum_{n\geq 1}\sum_{k\geq 1}\frac{1}{k\pi^{2k}(nm)^{2k}}\tag{2}$$ and so, summing over $m$ too: $$\sum_{m\geq 1}\log\left(m\sin\frac{1}{m}\right)=-\sum_{k\geq 1}\frac{1}{k\pi^{2k}}\sum_{s\geq 1}\frac{d(s)}{s^{2k}}=\color{red}{-\sum_{k\geq 1}\frac{\zeta(2k)^2}{k \pi^{2k}}}\tag{3}$$ that converges quite fast.

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Hint: Compute $$\lim_{x\to0}\frac{\log(x^{-1}\sin x)}{x^2}.$$

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  • $\begingroup$ Thank you. I have edited my answer to show what I attempted using the power series of $\sin$. $\endgroup$ – MathMan Jan 25 '15 at 21:55
  • $\begingroup$ Could you please have a look? $\endgroup$ – MathMan Jan 25 '15 at 21:57
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Since $$\log\left(n\sin \frac{1}{n}\right) = \log\left(n\left(\frac{1}{n} + O\left(\frac{1}{n^3}\right)\right)\right) = \log\left(1 + O\left(\frac{1}{n^2}\right)\right) = O\left(\frac{1}{n^2}\right)$$

and $\sum_{n = 1}^\infty \frac{1}{n^2}$ converges, so does $\sum_{n = 1}^\infty \log(n\sin\frac{1}{n})$.

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  • $\begingroup$ Thank you. I got it now :) $\endgroup$ – MathMan Jan 25 '15 at 22:20
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Hint: Show that $\log ({\sin x \over x})$ has a Taylor expansion of the form $cx^2 + $ higher order terms, and use $x = {1 \over n}$.

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  • $\begingroup$ Thanks. so, do I express it as $\log (1 + (\frac {\sin x}{x} - 1) )$ and then expand? $\endgroup$ – MathMan Jan 25 '15 at 22:03
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    $\begingroup$ Yes.. using the Taylor expansion of ${\sin x \over x}$. $\endgroup$ – Zarrax Jan 25 '15 at 22:10

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