For this series: $$\sum_{n=1}^{\infty}(-1)^n \frac{e^{x^2}+\sqrt{n}}{n^{\frac{3}{2}}},$$ I have no idea how to prove it converges uniformly in $[a,b]$. Can't even tell what does it converge to. I can just figure out that $$(-1)^n \frac{e^{x^2}+\sqrt{n}}{n^{\frac{3}{2}}}$$ converges to zero but have no idea what it is when it is summed up.
Can anybody help? Thanks a lot!
As per Gary's comment, first see that $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$ is just a convergent series of real numbers and has nothing to do with $x$. So you need not bother with that.
Secondly, $\displaystyle\sup_{x\in[a,b]}|\frac{(-1)^{n}e^{x^{2}}}{n\sqrt{n}}|\leq C|\frac{1}{n\sqrt{n}}|$ where $C=\max(e^{a^{2}},e^{b^{2}})$
And $\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}}$ converges. Thus, by Weirestrass M-test, the series $\sum_{n=1}^{\infty}\frac{(-1)^{n}e^{x^{2}}}{n\sqrt{n}}$ converges uniformly over $x\in[a,b]$ and that's it.
Warning/Caution: First you have to show the steps that I did above. Only then you can write $$\sum_{n=1}^{\infty}(-1)^n \frac{e^{x^2}+\sqrt{n}}{n^{\frac{3}{2}}}=\sum_{n=1}^{\infty}\frac{(-1)^{n}e^{x^{2}}}{n\sqrt{n}}+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$$ and then conclude uniform convergence not the otherway around!!!. If you are confused by this, then perhaps it is wise to ponder over it for a while.