Uniform convergence of $\sum\limits_{n=1}^{\infty} (-1)^n \frac{\arctan(x^n)}{\sqrt{n+1}}$ , $x \in[0,\infty)$

Question

I have a problem with checking the uniform convergence of

$\sum\limits_{n=1}^{\infty} (-1)^n \frac{\arctan(x^n)}{\sqrt{n+1}}$ , $x \in[0,\infty)$.

I suppose that using Abel-Dirichlet criterion should help, but I only know, that $\sum\limits_{n=1}^{\infty} \frac{1}{\sqrt{n+1}}$ is monotonically decreasing, non negative and bounded. And that the function $\frac{1}{\sqrt{n+1}}$ converges locally uniformly to $0$ on $(0,\infty)$ (I am also not sure, whether locally uniform convergence is enough for Dirichlet test, or if the uniform convergence on given interval is necessary) . But I am not able to show, that $\sum\limits_{n=1}^{\infty}(-1)^n \arctan(x^n)$ is uniformly convergent on $[0,\infty)$ (Abel) or that it has uniformly bounded partial sums (Dirichlet).

Or should I use completely different approach?

Thanks for any help.

Answer 1

It is not hard to show the pointwise convergence of the series, so I'm going to skip this step. The convergence allows us to rewrite the alternating series as follows:

$$\sum_{n=1}^\infty(-1)^n\frac{\arctan x^n}{\sqrt{n+1}}=-\frac{\arctan x}{\sqrt{2}}+\sum_{k=1}^{\infty}\arctan x^{2k}\left(\frac{1}{\sqrt{2k+1}}-\frac{1}{\sqrt{2k+2}}\right)+\sum_{k=1}^{\infty}\frac{\arctan x^{2k}-\arctan x^{2k+1}}{\sqrt{2k+2}}$$

The first two terms sum to a uniformly convergent series for all $x>0$, which can be proven using the Weierstrass M-test, with the estimate $|\arctan x|<\pi/2$. The second term is a bit trickier to work with. Using the identity for the tangent of the difference we find that

$\Delta_k(x):=\arctan x^{2k}-\arctan x^{2k+1}=\arctan\left(x^{2k}\frac{1-x}{1+x^{4k+1}}\right)$

Note that for any $k$, $\lim_{x\to\{0,1,\infty\}}\Delta_k(x)=0$. We must obtain an estimate on the supremum of this function in the intervals $(0,1)$ and $(1,\infty) $ separately, however the supremum is difficult to access analytically.

• If $0\leq x<1$ we note that

$$|\Delta_k(x)|\leq\arctan x^{2k}(1-x)\Rightarrow \sup_x |\Delta_k(x)|\leq\left(\frac{2k}{2k+1}\right)^{2k}\frac{1}{2k+1}\leq \frac{1}{2ke}$$

• For $1\leq x<\infty$ we estimate similarly

$$|\Delta_k(x)|\leq\arctan x^{-2k-1}(x-1)\Rightarrow \sup_x |\Delta_{k}(x)|\leq \frac{1}{2ke} $$

which in hindsight should have been expected due to the property $\Delta_k(x)=-\Delta_k(1/x)$. This bound is independent of $x$ and allows us to apply the Weierstrass M-test to show that the second series is also uniformly convergent. This guarantees that the sum of the two is also uniformly convergent to the right answer and in fact, we have surreptitiously proven that

$$\left|\sum_{n=N}^\infty(-1)^n\frac{\arctan x^n}{\sqrt{n+1}}\right|=\mathcal{O}\left(\frac{1}{N^{1/2}}\right)$$