Show that $\phi(x):=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}(1+\frac{x^{2}}{n})^{n}}$ is differentiable on $\mathbb{R}$.

Note that $2x/\sqrt{n}$ is monotonically decreasing with respect to $n$ and uniformly convergent to $0$ on $[0,R]$, We also have $-2x/\sqrt{n}$ montonically decreasing with respect to $n$ and uniformly convergent to $0$ on $[-R,0)$. Hence, the series

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

converges uniformly by Dirichlet's test on both $[-R,0)$ and $[0,R]$, and, therefore, on $[-R,R]$.

The sequence $(1 + x^2/n)^{-(n+1)}$ is eventually monotone and uniformly bounded for $x \in [-R,R]$. Therefore, by Abel's test we have uniform convergence of

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

Knowing that $\sum_{n=1}^{\infty} u_n(x)$ converges locally uniformly, here is a quick proof: Write

\begin{align*} u_n'(x) = -\frac{2x}{1+\frac{x^2}{n}} u_n(x) = -2x u_n(x) + \frac{2x^3}{n+x^2} u_n(x). \end{align*}

Then on each interval $[-R, R]$,

• $\sum_{n=1}^{\infty} (-2x) u_n(x)$ converges uniformly, and

• $\left| \frac{2x^3}{n+x^2} u_n(x) \right| \leq \frac{2R^3}{n^{3/2}} $ uniformly in $n$ and $x$, and so, $\sum_{n=1}^{\infty} \frac{2x^3}{n+x^2} u_n(x)$ converges uniformly by the Weierstrass M-test.

Therefore $\sum_{n=1}^{\infty} u_n'(x)$ also converges uniformly on $[-R, R]$.

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Addendum. Another purpose of this answer is to hint a much more general idea: You may split the sum into two parts: conditionally convergent part (which is easier to control) and absolutely convergent part.

In OP's case, we easily check that, for each $R> 0$,

$$ \frac{1}{(1+\frac{x^2}{n})^{n+1}} = e^{-x^2} + \mathcal{O}\left(\frac{1}{n}\right) $$

uniformly in $n$ and $x \in [-R, R]$, where the implicit bound of $\mathcal{O}(\frac{1}{n})$ depends only on $R$. Using this, we may write

$$ u_n'(x) = \frac{(-1)^{n+1}2xe^{-x^2}}{\sqrt{n}} + \mathcal{O}\left(\frac{1}{n^{3/2}}\right), $$

which can be used to easily prove the uniform convergence of $\sum_{n=1}^{\infty} u_n'(x)$ over $[-R, R]$.