Some help with the uniform convergence of the next series:
a) $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}sin(1+\frac{n}{x})$ for $x\ne 0$
b) $\sum_{n=1}^\infty \frac{sin(nx)}{n}e^{-nx}$ for $x>0$
c) $\sum_{n=1}^{\infty}nx^n$ for $0\leq x<1$
I've tried the M-test in a) and b) but i've failed, and the last one i tried to rewrite as quotient in the form $\frac{1-x}{1-x^m}$ for some m, but i also failed. Please I need understand how to prove or disprove the UC of these series, anyone can give a hand for this?
You can start the third one in this way:
$\sum_{n=1}^{\infty}x^n=\frac{x}{1-x}$
$\sum_{n=2}^{\infty}x^n=\frac{x}{1-x}-x=\frac{x^2}{1-x}$
$\sum_{n=3}^{\infty}x^n=\frac{x^2}{1-x}-x^2=\frac{x^3}{1-x}$
In general...
$\sum_{n=k}^{\infty}x^n=\frac{x^{k-1}}{1-x}-x^{k-1}=\frac{x^k}{1-x}$
if you sum all the series on the left you have: $x+2x^2+3x^3+\cdots+kx^k=\sum_{n=1}^{k}nx^n$ and, in the other side you have $\frac{x}{1-x}+\frac{x^2}{1-x}+\cdots \frac{x^3}{1-x}=\frac{1}{1-x}\sum_{n=1}^{k}x^k$. Now, taking $\lim_{k\rightarrow\infty}$ in both expressions, conclude.