Disprove uniform convergence of $\sum_{n=1}^{\infty} \frac{x}{(1+x)^n}$ in $[0,\infty)$ How would I show that $\sum_{n=1}^{\infty} \frac{x}{(1+x)^n}$ does not uniformly converge in $[0,\infty)$? 
I don't know how to approach this problem.
Thank you.

This is almost the same as Davide' answer:
let 
$$f_n(x)={x\over (1+x)^n},\ n\in\Bbb N^+;\ \ \text{ and }\ \ f(x)= \sum\limits_{n=1}^\infty {x\over(1+x)^n}.$$  Since, for $x>0$, the series $\sum\limits_{n=1}^\infty {1\over(1+x)^n}$ is a  Geometric series with $r={1\over 1+x}$:
$$
 f(x)=x\sum_{n=1}^\infty {1\over(1+x)^n}  =x\cdot{ 1/(1+x)\over 1-\bigl(1/(1+x)\bigr)}
=x\cdot{1\over x}=1,
$$
for $x>0$. 
As  $f(0)=0$, we see that  $f(x)$ converges pointwise  to a discontinuous function on $[0,\infty)$. Since a uniform limit of a sum of continuous functions is continuous, and as each term $f_n$ is continuous on $[0,\infty)$, it follows that $f(x)$ does not converge uniformly on $[0,\infty)$.
