# Cauchy criterion for sequences convergence

The task is to find out if the following sequence converges: $x_n = 1 + \dfrac{\sin(1)}{1^2} + \dfrac{\sin(2)}{2^2} + \ldots + \dfrac{\sin(n)}{n^2}$
I don't even know what to do, can you help me in any way?
(I know the Cauchy criterion, but can't figure out how to use it in this task)

• Convergent $\implies$ Cauchy. Can you bound this by another convergent series? – Rellek Dec 24 '17 at 14:39
• Yes, sines can be replaced with 1's, but it gives nothing – mr.cipriani Dec 24 '17 at 14:40
• Is $\sum_{n=1}^\infty \frac{1}{n^2}$ convergent? – Rellek Dec 24 '17 at 14:41
• Does integral test ring a bell? Or perhaps Cauchy condensation test? – Rellek Dec 24 '17 at 14:43
• $$\zeta(2)=\sum_{n\geq 1}\frac{1}{n^2} = 1+\sum_{n\geq 1}\frac{1}{(n+1)^2} \leq 1+\sum_{n\geq 1}\frac{1}{n(n+1)} = 2.$$ – Jack D'Aurizio Dec 24 '17 at 14:46

$$\sum_{n\geq 1}\frac{\sin(n)}{n^2} = \text{Im}\,\text{Li}_2(e^i)=-\text{Im}\int_{0}^{e^i}\log(1-x)\frac{dx}{x}=-\text{Im}\int_{1}^{e^i}\log(1-x)\frac{dx}{x}$$ equals $$-\text{Re}\int_{0}^{1}\log(1-e^{ix})\,dx=\int_{0}^{1}-\log\left(2\sin\frac{x}{2}\right)\,dx$$ which is well approximated by $\int_{0}^{1}-\log(x)\,dx = 1.$ Indeed $$2\sin\frac{x}{2} = x\prod_{n\geq 1}\left(1-\frac{x^2}{4n^2\pi^2}\right)$$ so $$\sum_{n\geq 1}\frac{\sin(n)}{n^2} = 1+\sum_{m\geq 1}\frac{\zeta(2m)}{4^m \pi^{2m} m(2m+1)}$$ and an even better approximation of the LHS is $1+\frac{2}{141+\sqrt{5}}$.

So you know that this series is bounded above by the series with terms $\frac{1}{n^2}$ since $|\sin(x)| \leq 1$. Now notice that $$\frac{1}{n(n+1)} \leq \frac{1}{n^2} \leq \frac{1}{n(n-1)}$$ We also know that $$\sum_{n=1}^{m} \frac{1}{n(n+1)} = \sum_{n=1}^{m} (\frac{1}{n} - \frac{1}{n+1}) = 1-\frac{1}{m+1}$$ Now, this converges to $1$. We also know that $\sum \frac{1}{n^2}$ is of the same nature of $\sum \frac{1}{n(n+1)}$. What can you deduce?

Let $S_n =$

$1+ \sin(1)/1^1 + \sin(2)/2^2....$

$+\sin(n)/n^2$.

$|S_n|$

$\le |1|+|\sin(1)/1^2| +.......$

$|sin(n)/n^2| \le \sum_{k=1}^{n} \dfrac {1}{n^2} .$

$\sum_{k=1}^{n}\dfrac{1}{n^2}$ converges.

The series $S_n$ is absolutely convergent, the partial sums of the absolute values of the terms are mon. increasing and bounded above, hence convergent.

$\rightarrow :$

$S_n$ converges.

Note:

$\sum_{k=2}^{n}\dfrac{1}{k^2} \lt \sum_{k=1}^{n}\dfrac{1}{k(k+1)} =$

$\sum_{k=1}^{n}[\dfrac{1}{k} - \dfrac{1}{k+1}] = 1- \dfrac{1}{n+1};$

Hence bounded above , mon increasing:

$\rightarrow:$

$\sum \dfrac{1}{n^2}$convergent.

• OP wasn't sure why $\sum\frac1{n^2}$ converges, so perhaps it would be useful to add this to the answer? – John Doe Dec 24 '17 at 14:59
• John.What does OP stand for? – Peter Szilas Dec 24 '17 at 15:20
• "Original poster" - so basically the person that asked the question - mr.cipriani – John Doe Dec 24 '17 at 15:21
• Thanks, John D. – Peter Szilas Dec 24 '17 at 16:21