# Uniform convergence of the series $\sum_{n=1}^\infty\frac{x}{(1+x)^n}$

I have to determine whether or not the series $$\sum_{n=1}^\infty\frac{x}{(1+x)^n}$$ converges uniformly on $$[0,1]$$.

I attempted to use Weierstrass' M-Test, by finding the maxima of each $$a_n$$ which must exist by the maximum value theorem. The maxima turned out to be at $$\dfrac{1}{n-1}$$, and to be equal to $$\dfrac{(n-1)^{n-1}}{n^n}$$. The sum of these terms diverges, and hence I can't use Weierstrass' M-Test.

Now, I'm thinking of using Dini's Theorem since $$[0,1]$$ is compact, and seting $$f_n(x)$$ to be the $$n$$-th partial sum gives us a monotonic increasing sequence. So if I can find the pointiwse limit of this sequence, and verify it is continuous, I will be done.

Is this the right track? If it is, how would I compute the series?

Note that as $$N\to \infty$$ $$\sum_{n=1}^{N}\dfrac{x}{(1+x)^n}=1-\frac{1}{(1+x)^{N}}\to \begin{cases}0&\text{if x=0,}\\ 1&\text{if x>0.}\end{cases}$$ So the pointwise limit function is not a continuous in $$[0,a)$$ with $$a>0$$. What may we conclude about the uniform convergence over such interval? What happens over $$[a,b]$$ with $$0?

• I see, so there is no uniform convergence. But how do we obtain that expansion? Aug 4, 2019 at 13:21
• The series is geometric with $r=1/(1+x)$. See en.wikipedia.org/wiki/Geometric_series#Formula Aug 4, 2019 at 13:25

For $$0, the series converges to $$\frac{x}{1+x}\frac1{1-\frac1{1+x}}=1$$ (geometric series). For $$x=0$$ it converges to $$0$$. Is the limit continuous?