6
$\begingroup$

Could you give me some hint how to deal with this series ?

I could not conclude about absolute convergence because $\frac {\left|\sin\left( n+\frac 1n \right)\right|}{(\ln n)^2}\le \frac 1{(\ln n)^2}$ does not get me anywhere.

I tried to use the Dirichlet's test to prove convergence of this series: $\sum_{n=2}^\infty \frac {\sin\left( n+\frac 1n \right)}{(\ln n)^2}$ but I could not prove that $\sum_{n=2}^\infty \sin(n+\frac 1n)$ are bounded.

Thanks.

$\endgroup$
3
$\begingroup$

$\sin (n+\frac{1}{n}) = \sin n \cos \frac{1}{n} + \sin \frac{1}{n} \cos n$

Firstly $\sum_{n=2}^\infty \frac{\sin \frac{1}{n} \cos n}{\ln ^2 n}$ is absolutely convergent from comparison test, because

$\sum_{n=2}^\infty \left| \frac{\sin \frac{1}{n} \cos n}{\ln ^2 n} \right| \le \sum_{n=2}^\infty \frac{\frac{1}{n} \cdot 1}{\ln ^2 n} < \infty$

Convergence of last series follows for example from http://en.wikipedia.org/wiki/Cauchy%27s_condensation_test

Secondly observe, that function $f(x)=\frac{\cos \frac{1}{x}}{\ln ^2 x}$ is decraising for big $x$ because $f'(x)= \frac{\ln x \sin \left( \frac{1}{x} \right) -2x \cos \left( \frac{1}{x} \right)}{\ln ^3 x}$,

Easy to see that $\lim_{x\to \infty} f'(x) = -\infty$ therefore there is $M>0$ such that for $\mathbb{N} \ni n> M$ sequence $\left( f(n) \right)_{n>M}$ is decraising and tends to $0$. If we could show that $\left|\sum_{k=1}^n \sin k \right| $ is bounded, we would just apply Dirichlet test to show that $\sum_{n=2}^\infty \frac{\sin n \cdot \cos \frac{1}{n}}{\ln ^2 n}$ diverges.

It's well known that $\sum_{k=1}^n \sin ka = \frac{\sin \left(\frac{na}{2} \right) \sin \left( \frac{(n+1)a}{2} \right)}{\sin \left( \frac{a}{2}\right)}$ (you can prove it f.e by complex numbers), so

$\left| \sum_{k=1}^n \sin k \right| \le \frac{1}{\left| \sin \frac{1}{2}\right|}$.

To sum up: $\sum_{n=2}^\infty \frac{\sin (n+ \frac{1}{n})}{\ln ^2 n}$ converges as sum of two convergent series.

$\endgroup$
3
$\begingroup$
  1. The series $\sum_{n\geqslant 2}\frac{\sin n}{(\log n)^2}$ is convergent (use a summation by parts argument).

  2. We have $\left|\sin\left(n+\frac 1n\right)-\sin n\right|\leqslant \frac 1n$.

  3. The series $\sum_{n\geqslant 2}\frac 1{n(\log n)^2}$ is convergent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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