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

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 $\log$.

• dfrac and displaystyle are not appropriate for titles you can find more about this on meta. – dustin Jan 25 '15 at 21:45
• @Wanderer $\lim_{n\to \infty} n\sin \frac{1}{n} = 1$, so $\lim_{n\to \infty} \log(n\sin \frac{1}{n}) = 0$. – kobe Jan 25 '15 at 21:47
• @kobe Could you please explain further. I mean, I have already used this result above. – MathMan Jan 25 '15 at 21:48
• 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 $\sin x$? – Harald Hanche-Olsen Jan 25 '15 at 21:49
• @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 $$\lim_{n\to \infty}\log(n\sin\frac{1}{n}) = \log(\lim_{n\to \infty} n\sin \frac{1}{n}) = \log(1) = 0$$ – kobe Jan 25 '15 at 21:50

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.

• As a side note, if you are sufficiently versed in these expansions, then you can do this heuristic quickly and in your head. – davidlowryduda Jan 25 '15 at 22:06
• Minor nit: Where you write (error $\ll\frac1{x^4}$), the $\ll$ is a bit too strong. – Harald Hanche-Olsen Jan 26 '15 at 10:57

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.

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

• Thank you. I have edited my answer to show what I attempted using the power series of $\sin$. – MathMan Jan 25 '15 at 21:55
• Could you please have a look? – MathMan Jan 25 '15 at 21:57

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 $\sum_{n = 1}^\infty \log(n\sin\frac{1}{n})$.

• Thank you. I got it now :) – MathMan Jan 25 '15 at 22:20

Hint: Show that $\log ({\sin x \over x})$ has a Taylor expansion of the form $cx^2 +$ higher order terms, and use $x = {1 \over n}$.

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