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.

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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 $[0,+\infty)$.

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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 $[0,\infty)$.

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$$\sup_{x \in [0,\infty)} \sum_{n=m}^{2m} \frac{x}{(1+x)^n} \geq \sup_{x \in [0,\infty)} \frac{mx}{(1+x)^{2m}} \geq \frac{mx}{(1+x)^{2m}} \bigg|_{x=1/m} = \frac1{(1+ \frac{1}{m} )^{2m}}\to \frac1{e^2}.$$

Thus the series is not uniformly Cauchy on $[0,\infty).$

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It seems the denominators $(1+x)^m$ should read $(1+x)^{2m}$ (with the adequate subsequent modifications, which fortunately preserve the positivity of the limit). – Did Feb 26 '12 at 10:50