The series is:
$$\sum_{n=1}^\infty \frac{\cos(\sqrt{n})}{n}$$
Considering it isn't always positive, I replace $\frac{\cos{\sqrt{n}}}{n}$ with its absolute value and I find that:
$$\vert \frac{\cos{\sqrt{n}}}{n}\vert\gt \frac{\cos^2{\sqrt{n}}}{n}=\frac{\ 1+\cos{2\sqrt{n}}}{2n}=\frac{1}{2n}+\frac{\cos{2\sqrt{n}}}{2n}$$
if $\sum_{n=1}^\infty \vert\frac{\cos{\sqrt{n}}}{n}\vert $ converges, then
$\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$ converges.Using Comparison test,we can draw the conclusion that $\sum_{n=1}^\infty\frac{1}{2n}$ converges , which is impossible.
So I get that $\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$ absolutely diverges. But I can't figure out whether $\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$ converges or not.
I have tried Dirichlet's test, but I can't figure out whether $$S_{n}=\sum_{k=1}^n \cos{\sqrt{k}}$$ is bounded.
(This is my first time to ask question.Maybe there exist some mistakes in my conclusion.Thanks. 🙂
Best Answer
We show that $N\to\sum_{n = 1}^N {\frac{\cos(\sqrt{n})}{n}}$ is a Cauchy sequence and therefore the given series is convergent.
Hint. By using the MVT prove that for $n\geq 1$ and for all $x\in [n,n+1)$ $$\left|\frac{\cos(\sqrt{n})}{n}-\frac{\cos(\sqrt{x})}{x}\right|\leq \frac{1}{n^{3/2}}.$$ Then for $M>N\geq 1$, $$\begin{align} &\left|\sum_{n = N}^M {\frac{\cos(\sqrt{n})}{n}} - \int_N^{M + 1} {\frac{{\cos (\sqrt{x} )}}{x}dx} \right|\\ &=\left|\sum_{n = N}^M {\frac{\cos(\sqrt{n})}{n}} - \sum_{n = N}^M {\int_n^{n + 1} {\frac{{\cos (\sqrt{x} )}}{x}dx} } \right|\\ &\leq\sum_{n = N}^M \int_n^{n + 1}\left|{\frac{\cos(\sqrt{n})}{n}} - { {\frac{{\cos (\sqrt{x} )}}{x}} } \right|dx \le \sum_{n = N}^M \frac{1}{n^{3/2}}.\end{align}$$ Note that since $\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ is convergent then $$\lim_{M,N\to +\infty} \sum_{n = N}^M \frac{1}{n^{3/2}}=0.$$ Moreover $$\int_1^{+\infty}\frac{{\cos (\sqrt{x} )}}{x}dx=2\int_1^{+\infty}\frac{{\cos (u )}}{u}\, du$$ where the last integral is convergent, and we have that $$\lim_{M,N\to +\infty}\int_N^{M+1}\frac{{\cos (\sqrt{x} )}}{x}dx=0.$$