I'm trying to solve this problem about series of functions:
Study the pointwise and uniform convergence of the following series:
$$\sum_{n=1}^\infty \left(\frac{\sqrt n – \sin(x)}{n^2x^n}\right)^2$$
I studied the pointwise convergence and I got that the series converges pointwise in
$$(-\infty,-1] , [1,+\infty)$$
Then I tried to study the total convergence in $[1,+\infty)$ using the M-test :
$$\left(\frac{\sqrt n +(- \sin(x))}{n^2x^n}\right)^2\le\left(\frac{\sqrt n + 1}{n^2x^n}\right)^2<\left(\frac{\sqrt n+1}{n^2}\right)^2=M(n)$$
Since $\sum_{n=1}^\infty M(n) <+\infty$ the series converges totally in $[1,+\infty)$ (I am not really sure about this).
When I try to study the total convergence in $(-\infty,-1]$ I can't get to the same result using the M-test. I tried using the Weiestrass test but studying the first derivative gets me to an inequality involving sin cos and x, and I'm not sure I can solve it. What should I do to solve this problem? Thanks, and sorry if my english is not perfect.
Best Answer
For any $x\ne 0$ and $n\ge1$, we have
$$\begin{align} \left|\left(\frac{\sqrt n-\sin(x)}{n^2x^n}\right)^2\right|&\le \frac{(\sqrt n+1)^2}{n^4x^{2n}}\\\\ &\le \frac{4}{n^3 x^{2n}} \end{align}$$
And for $|x|\ge 1$, we find that
$$\left|\left(\frac{\sqrt n-\sin(x)}{n^2x^n}\right)^2\right|\le \frac{4}{n^3}$$
Now applying the M-test, we find that for $|x|\ge 1$, the series of interest converges uniformly.