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Prove that $\displaystyle \sum_{n=1}^\infty \;\ln\left(n\sin\displaystyle \frac{1}{n}\right)$ converges.

My Work:

$\Biggl|\ln\Biggl(n\sin\displaystyle \frac{1}{n}\Biggr)\Biggr|\leq \Biggl|\ln\Biggl(n\sin\displaystyle \frac{1}{n^2}\Biggr)\Biggr|\leq \Biggl|\ln\Biggl(\sin\displaystyle \frac{1}{n^2}\Biggr)\Biggr|$. I was going to use comparison test. But now stuck. Please give me a hint.

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Hint: try to use the fact that $$ \ln\left(n\sin \frac{1}{n}\right) = \ln\left(n \left( \frac{1}{n}+O\left(\frac{1}{n^3}\right) \right)\right) = \ln\left( 1+O\left(\frac{1}{n^2} \right)\right) = O\left(\frac{1}{n^2} \right) $$ and the comparison test.

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  • $\begingroup$ Can you plese elaborate the last step? $\endgroup$ – Extremal Jan 3 '15 at 4:28
  • $\begingroup$ Did you use taylor series for ln? $\endgroup$ – Extremal Jan 3 '15 at 4:29
  • $\begingroup$ Just the equivalent $\ln(1+x)\sim_{x\to 0} x$; a Taylor series (with more terms) would also work. $\endgroup$ – Clement C. Jan 3 '15 at 13:12
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Let's amply use Taylor series to give the leading orders.

Note that $\sin\left( \frac{1}{n} \right) \approx \frac{1}{n} - \frac{1}{6n^3}$. So $n \sin \left( \frac{1}{n} \right) \approx 1 - \frac{1}{6n^2}$.

Note also that $\ln(1 - x) \approx x + x^2 + \dots$, so that $$\ln\left( n \sin \frac{1}{n} \right) \approx \frac{1}{6n^2} + \frac{1}{36n^4} + \dots$$

This means that you are wondering about $$ \sum_{n \geq 1} \sum_{j \geq 1} \frac{1}{(6n^2)^j} \ll \sum_{n \geq 1} \frac{1}{n^2},$$ which converges.

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  • $\begingroup$ I think that you have some typo's in the signs. Cheers. $\endgroup$ – Claude Leibovici Jan 3 '15 at 7:00
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In the same spirit as mixedmath's answer, let us massively use Taylor expansions for large values of $n$. Then $$\sin(\frac 1n)=\frac{1}{n}-\frac{1}{6 n^3}+\frac{1}{120 n^5}-\frac{1}{5040 n^7}+O\left(\left(\frac{1}{n}\right)^8\right)$$ $$n\sin(\frac 1n)=1-\frac{1}{6 n^2}+\frac{1}{120 n^4}-\frac{1}{5040 n^6}+O\left(\left(\frac{1}{n}\right)^8\right)$$ $$\log \left(n \sin \left(\frac{1}{n}\right)\right)=-\frac{1}{6 n^2}-\frac{1}{180 n^4}-\frac{1}{2835 n^6}+O\left(\left(\frac{1}{n}\right)^8\right)$$ Summing from $n=1$ to $\infty$, then $$S=\sum_{n=1}^\infty \ln\big(n\sin \frac{1}{n}\big)\approx -\frac 16 \sum_{n=1}^\infty\frac{1}{n^2}-\frac 1{180} \sum_{n=1}^\infty\frac{1}{n^4} -\frac 1{2835} \sum_{n=1}^\infty\frac{1}{n^6}$$ and the value of the successive sums are $\frac{\pi ^2}{6}$, $\frac{\pi ^4}{90}$, $\frac{\pi ^6}{945}$ which make $$S \approx -\frac{\pi ^2 \left(595350+1323 \pi ^2+8 \pi ^4\right)}{21432600}\approx -0.280527$$ while the numerical evaluation of the original sum is $-0.280556$.

Adding an extra term to the expansion would lead to $S \approx -0.280554$.

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  • $\begingroup$ Sorry for late comment, but where do you get the -1/180 from? $\endgroup$ – Quality Mar 1 '15 at 1:04

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