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.
 A: We can write $\frac x{(1+x)^n}=\frac {x+1-1}{(1+x)^n}=\frac 1{(1+x)^{n-1}}-\frac 1{(1+x)^n}$ so the partial sum is $1-\frac 1{(1+x)^n}$. We only have to look at uniform convergence on $[0,+\infty)$ of $f_n(x):=\frac 1{(1+x)^n}$. It converges pointwise to the function $x\mapsto \begin{cases}1&\mbox{ if }x=0\\\ 0 &\mbox{ if }x>0\end{cases}$, which is not continuous. So the convergence cannot be uniform on $[0,+\infty)$. 
A: 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 $[0,\infty)$.
A: $$ \sup_{x \in [0,\infty)} \sum_{n=m}^{2m} \frac{x}{(1+x)^n}  \geq \sup_{x \in [0,\infty)}  \frac{mx}{(1+x)^{2m}}  \geq  \frac{mx}{(1+x)^{2m}} \bigg|_{x=1/m} = \frac1{(1+ \frac{1}{m} )^{2m}}\to \frac1{e^2}.$$
Thus the series is not uniformly Cauchy on $[0,\infty).$
