# I need to show that functional series $\sum_{n=1}^{+\infty}\frac{\sqrt{n+x^2}}{n^2+e^x}$ is differentiable

I have a function sequence $$f_n(x) = \frac{\sqrt{n+x^2}}{n^2+e^x},\quad x \in \mathbb{R}$$

I define a function series $$F(x):=\sum_{n=1}^{+\infty}f_n(x), \quad F:\mathbb{R} \to \mathbb{R}$$

I neet to show:
$1)$ that $F$ is differentiable.
$2)$ that $\lim_{x \to 0}F = g$ is exists(over) and if exists that $g > \frac{3}{4}$

$1)$ So I think that
$$f_n'(x) = \frac{x}{(n^2+e^x)\sqrt{n+x^2}} - \frac{e^x\sqrt{n+x^2}}{(n^2+e^x)^2}$$ and if I show that $f_n'(x)$ is uniform convergence, then I have that F is differentiable($f_n$ is pointwise convergence), but I don't now how to show it.
2) As for the second, I don't know how to approach this.

All the functions being summed are positive, and

$$\frac{\sqrt{n+x^2}}{n^2+e^x}\le\frac{\sqrt{2n}}{n^2}=\frac{\sqrt2}{n^{3/2}}\implies \text{ Since the positive series}\;\;\sum_{n=1}^\infty\frac{\sqrt2}{n^{3/2}}\;\;\text{converges,}$$

Weierstrass M-test gives us absolute and uniform convergence in compact subsets of the original series for $\;x\in\Bbb R\;$ .

Also observe that for any $\;n\in\Bbb N\;$ :

$$f'_n(x)=\frac{\frac x{\sqrt{n+x^2}}(n^2+e^x)-e^x\sqrt{n+x^2}}{(n^2+e^x)^2}=\frac{xn^2+(x-n-x^2)e^x}{\sqrt{n+x^2}\,(n^2+e^x)^2}$$

Show that also the series of the above derivatives converges absolutely and uniformly and thus the limit function is differentiable and its derivative is the series sum of the derivatives.

• but for which $?$ $f_n'(x) \leq ?$ is true? – A. Miller Mar 19 '18 at 10:28
• @A.Rokinsky For all of them, as far as I can see:$$\left|\frac{xn^2+(x-n-x^2)e^x}{\sqrt{n+x^2}\,(n^2+e^x)^2}\right|\le\frac{|x|n^2}{\sqrt n\,n^4}\ldots...$$ – DonAntonio Mar 19 '18 at 10:33