Does the series: $\displaystyle\sum^\infty_{n=1}\frac 1{\sqrt n} \sin\left(\frac1n\right)$ converge?
I think we can do this: $\frac 1 {\sqrt n}\sin(\frac 1 n) \sim \frac 1 {\sqrt n}\frac 1 n=\frac 1 {n^{1.5}}$ and we know that its converging since the exponenet is $>1$.
But when I'm trying the limit comparison test with $\frac 1 {\sqrt n}$ I get that the series actually diverges like $\frac 1 {\sqrt n}$ does…
So is there another way to do this without the $\sim$ I did above?
Note: no integral test.
Best Answer
Simply:
$$0\leq \sum_{n\geq 1}\frac{1}{\sqrt{n}}\,\sin\frac{1}{n}\leq \sum_{n\geq 1}\frac{1}{n\sqrt{n}}=\zeta(3/2).$$ To provide an explicit bound for $\zeta(3/2)$, notice that: $$\frac{1}{n\sqrt{n}}\leq\frac{2}{\sqrt{n-\frac{1}{2}}}-\frac{2}{\sqrt{n+\frac{1}{2}}}$$ implies, by telescopic summation, $$\sum_{n\geq 1}\frac{1}{n\sqrt{n}} \leq \sqrt{8}.$$