Show that the series converges to a function $S(x)\in C^1$.

Question

Let $$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt n} \arctan(\frac{x}{\sqrt n})$$. Show that the function converges uniformly for every $x\in\mathbb{R}$ to a function, $S(x)$ which is in $C^1$.

So I thought about looking at the series of the derivatives:

$$\sum_{n=1}^\infty \frac{(-1)^n}{n}\frac{1}{1+\frac{x^2}{n}}$$

Now, I expected to get a series which converges uniformly on $\mathbb{R}$ by using Weirestarss M-test, but that doesn't seem the case since $\sum \frac{1}{n} = \infty$.

What should I do?

Answer 1

Show that for all $\epsilon\gt 0$, there exists an $N$ such that, when $n \gt N$,

$$\left |\arctan{\frac{x}{\sqrt{n}}} - \frac{x}{\sqrt{n}} \right | \lt \epsilon \left (\frac{x}{\sqrt{n}} \right )^3$$

such that the series may be split into a finite part and

$$x \sum_{N+1}^{\infty} \frac{(-1)^n}{n} $$

with an error $O(\epsilon x^3)$. That second sum converges by comparison with the series for $\log{2}$.

Answer 2

First we can see that for $x_0=0$ the series converges to $0$. Next, we want to observe the $\sum_{n=1}^\infty f'_n(x)$ and check for uniform convergence.

$$\sum_{n=1}^\infty f'_n(x) = \sum_{n=1}^\infty \frac{(-1)^n}{n+x^2}$$

We can use Dirichlet's test:

• $a_n = \frac{1}{n+x^2}$ converges monotonically and uniformly to $0$.
• $\sum b_n = \sum (-1)^n$ is obviously bounded by $12$ for example.

So we have that $\sum_{n=1}^\infty f'_n(x)$ converges uniformly to some $S'(x)$ and therefore $\sum_{n=1}^\infty f_n(x)$ converges uniformly to some $S(x)$. Moreover, since $f'_n(x)$ is continuous then $S'(x)$ is also continuous (because the convergence is uniform) and therefore, $S(x)\in C^1$.