This question already has an answer here:

I must determine whether the following series converges: $$\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)$$

I know that in general, I must use the limit comparison test, but I cannot find an expression to which I can compare it. For instance, I have tried the usual process:

For $n$ large, we have that $\lim_{n\to \infty}n\sin\frac1n=1$, and so, $\ln(1)=0$. This fails the divergence test, but it cannot be concluded automatically that the series is convergent either. How may I proceed here? Any help would be appreciated.


marked as duplicate by YuiTo Cheng, Lord Shark the Unknown, postmortes, José Carlos Santos calculus Jun 25 at 12:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ Hint: What is the Taylor series of $\sin x$, and of $\ln(1+t)$ ? $\endgroup$ – Lucian Mar 15 '15 at 4:19
  • 1
    $\begingroup$ $n\sin(1/n)\approx 1-\frac{1}{6n^2}$ and $\log (1-x)\approx -x$, so I'd guess yes. $\endgroup$ – Thomas Andrews Mar 15 '15 at 4:20
  • 1
    $\begingroup$ By the way, saying "For large $n$, $\lim_{n\to\infty}\dots$" is redundant." $\endgroup$ – Thomas Andrews Mar 15 '15 at 4:22

Using Taylor series, $$\sin x\sim x-\frac{x^3}{3!}+o(x^3)$$$$\ln(1+x)\sim x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$$ When $n\rightarrow\infty$, $$n\sin\frac{1}{n}\sim n(\frac{1}{n}-\frac{1}{3!n^3}+o(\frac{1}{n^3}))=1-\frac{1}{6n^2}+o(\frac{1}{n^2})$$

Thus, $$\ln(n\sin\frac{1}{n})\sim -\frac{1}{6n^2}+o(\frac{1}{n^2})-\frac{\left(-\frac{1}{6n^2}+o(\frac{1}{n^2})\right)^2}{2}+o\left(-\frac{1}{6n^2}+o(\frac{1}{n^2}\right)=-\frac{1}{6n^2}+o(\frac{1}{n^2})$$

As $\sum\dfrac{1}{n^2}$ is convergent, so is $\sum\ln(n\sin\dfrac{1}{n})$.

Hope this can help you.

  • $\begingroup$ Nicely done Kevin. $\endgroup$ – Gabriel Romon Mar 15 '15 at 10:18
  • $\begingroup$ I think one only needs one term in the expansion of $\ln(1+x)$. $\endgroup$ – Jack Jul 7 '17 at 14:22
  • 2
    $\begingroup$ Shouldn't '$\sim$' be '$=$'? $\endgroup$ – user531232 Aug 12 '18 at 5:18

$$S=\sum_{n\geq 1}\left(1-n\sin\frac{1}{n}\right)$$ converges and so does your series by $0\leq \log(1+x)\leq x$ over $[0,1]$.
$S$ has a nice integral representation: the sine function is entire, hence

$$ S = \sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{(2m+1)! n^{2m}}=\sum_{m\geq 1}\frac{\zeta(2m)(-1)^{m+1}}{(2m+1)!}=2\int_{0}^{+\infty}\!\!\!\!\underbrace{\frac{\text{Ber}_2(x)}{e^{x^2/4}-1}}_{\text{gaussian-shaped}}dx $$ by the integral representation for the $\zeta$ function, where $\text{Ber}_2$ is a Kelvin function.
By the fast-convergent series representation and Leibniz' test we have $0.2652\leq S\leq 0.2654$.


Since $$ \sin h \sim h-\frac{h^3}6 ~~\text{and}~~~\lim_{x\to 0}\frac{\ln(x+1)}{x} = 1$$ we have, $$\lim_{n\to\infty} \color{red}{n^2}\ln\left(n\sin\left(\frac{1}{n}\right)\right)=\lim_{h\to 0} \frac{\ln\left(\frac{\sinh-h}{h}+1\right)}{h^2} \sim \lim_{h\to 0}\frac{1}6 \frac{\ln\left(1-\frac{h^2}6\right)}{\frac{-h^2}6} =\frac{1}6 $$ Then, the exist N such that for $n>N$ we have,

$$\left|\ln\left(n\sin\left(\frac{1}{n}\right)\right)\right|\le \frac{c}{n^2}$$

this prove the convergence of $$\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)$$

since, $$\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$$


A variant using the equivalent of $\ln x$ near $x=1$:

As this is a series with negative terms, we can use equivalents.

Note that $\;\lim\limits_{n\to\infty}n\sin\dfrac1n=1$, and remember that if $x\to 1$, $\:\ln x\sim x-1$. Thus, using Taylor's formula at order $3$, $$\ln n\sin\dfrac1n \sim_{n\to\infty}\:n\sin\frac1n-1=n\Bigl(\frac1n-\frac1{6n^3}+o\Bigl(\frac1{n^3}\Bigr)\Bigr)-1=-\frac1{6n^2}+o\Bigl(\frac1{n^2}\Bigr) \sim_{n\to\infty}-\frac1{6n^2}, $$ which is a convergent series.


Taylor series of $\sin x$ is $$\sin x = x - \frac{x^3}{3!} + O(x^5)$$ Hence $$\sin \frac{1}{n} = \frac{1}{n}- \frac{1}{6n^3} + O(\frac{1}{n^5})$$ So $$n\sin \frac{1}{n} = 1- \frac{1}{6n^2} + O(\frac{1}{n^4})$$ Taylor series of $\ln$ is $$\ln(1+x) = x + O(x^2)$$ For $x = -\frac{1}{6n^2} + O(\frac{1}{n^4})$, we get $$\ln(1 - \frac{1}{6n^2} + O(\frac{1}{n^4})) = -\frac{1}{6n^2} + O(\frac{1}{n^4})$$ Then you'd get a sum of convergent series since all those sums have $p > 1$ according to the p-series test.


Taylor's theorem tells us that $\sin x = x - \dfrac1{3!} x^3 + \dfrac1{5!} \cos(\xi) x^5$ for some $\zeta \in [0,x]$ whenever $x > 0$. Since $0 \le \cos \xi \le 1$ when $x = \dfrac1n$:

$$x - \frac1{3!} x^3 \le \sin x \le x - \frac1{3!} x^3 + \frac1{5!}x^5$$

Substituting $x = \dfrac1n$:

$$\frac1n - \frac1{3!} \frac1{n^3} \le \sin \frac1n \le \frac1n - \frac1{3!} \frac1{n^3} + \frac1{5!} \frac1{n^5}$$

Multiplying all sides by $n$:

$$1 - \frac1{3!} \frac1{n^2} \le n \sin \frac1n \le 1 - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4}$$

Applying Taylor's theorem on $\ln(1-x)$ when $x>0$: $$\ln(1-x) = - \frac 1 {1 - \xi} x$$ for some $\xi \in [0,x]$

Therefore: $$-\frac{x}{1-x} \le \ln(1-x) \le -x$$

That is: $$1-\frac{1}{1-x} \le \ln(1-x) \le -x$$

Combining the above two sections:

$$\ln \left(1 - \frac1{3!} \frac1{n^2}\right) \le \ln \left( n \sin \frac1n \right) \le \ln \left(1 - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4} \right)$$


$$\frac{-1}{6n^2-1} \le \ln \left( n \sin \frac1n \right) \le - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4}$$

Applying sandwich theorem:

$$\sum_{n=1}^\infty \frac{-1}{6n^2-1} \le \sum_{n=1}^\infty \ln \left( n \sin \frac1n \right) \le \sum_{n=1}^\infty - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4}$$

and both sums can be easily shown convergent.


Not the answer you're looking for? Browse other questions tagged or ask your own question.