convergency of a function series I've got the following function series:
$$ f(x) = \sum_{n = 0}^{\infty} \frac{e^{nx}-1}{2^ne^{nx}}$$
Is it punctually convergent and uniformly convergent?

What I have done is:
I've chosen the series $\frac{1}{2^n}$ which is bigger than the series given in the problem, and because of the theory of Weierstrass, I've said that as $\frac{1}{2^n}$ is convergent, the given series is also convergent punctually and uniformly.
But I'm not sure this is OK. If it's not, how can I do it?

I believe this can be put into the form of a geometric series s.t. 
$$f(x) = \sum_{n = 0}^{\infty} \bigg(\frac{1}{2}\bigg)^n - \sum_{n = 0}^{\infty} \bigg(\frac{1}{2e^x}\bigg)^n $$
The first term can be evaluated without thought via the equation for the geometric series: $\sum_\limits{n =0}^{\infty} ar^k = \frac{a}{1-r}$, $\forall |r|<1$, 
$$\implies f(x) = 2 - \sum_{n = 0}^{\infty} \bigg(\frac{1}{2e^x}\bigg)^n$$ 
The second term is also in the form of a geometric series, however it is only convergent if $\bigg|\frac{1}{2e^x} \bigg| < 1$. 
More work is still require, but can you take it from here?
