Convergence of $\sum \frac{\sqrt{n}-n\sin (\frac{1}{\sqrt{n}})}{n}$.

Question

Does the sum in the title converge? $\sum \frac{\sqrt{n}-n\sin (\frac{1}{\sqrt{n}})}{n} = \sum (\frac{1}{\sqrt{n}}-\sin (\frac{1}{\sqrt{n}})$), so one can use the integral criteria, which says the sum converges iff the integral $I = \int_1^{\infty} (\frac{1}{\sqrt{x}}-\sin (\frac{1}{\sqrt{x}}))$ converges. Heuristically, $I$ is the area between $y=x$ and $y=\sin(x)$ on the interval $(0;1)$ and this area is evidently (heuristically) finite.

But is there some more analytical (= rigorous, classical) approach? Some convergent series to compare it with, for example? I'm just trying to find some solution (or find out whether there exists one) which would be accessible to an average student after the basic calculus course.

Answer 1

It is not really clear to me what you are after, but maybe something like this, which is close to Maclaurin, but a pure estimate?

Integrating (from $0$ to $x$) the well-known inequality $$ \sin x\leq x,\quad x\geq 0. $$ we get $$ 1-\cos x=\int_0^x\sin t\,dt\leq \int_0^x t\,dt= x^2/2. $$ Repeating, $$ x-\sin x=\int_0^x 1-\cos t\,dt\leq \int_0^x t^2/2\,dt= x^3/6. $$ Thus, for $x\geq 0$, we have $$ 0\leq x-\sin x\leq x^3/6. $$ With $x=1/\sqrt{n}$, $$ 0\leq\frac{1}{\sqrt{n}}-\sin\frac{1}{\sqrt{n}}\leq\frac{1}{6n^{3/2}} $$ Since $$ \sum_{n=1}^{+\infty}\frac{1}{6n^{3/2}} $$ converges, so do $$ \sum_{n=1}^{+\infty}\Bigl(\frac{1}{\sqrt{n}}-\sin\frac{1}{\sqrt{n}}\Bigr). $$