Determine the convergence, both pointwise and uniform, on [0,1] for the series
$\displaystyle \sum_{n = 1}^{\infty} [e^{-nx}-e^{-(n+1)x}] $
hint: look at the partial sums!
Note that
$\displaystyle \displaystyle e^{-nx}-e^{-(n+1)x}=\left( \frac{1}{e^{x}}\right)^n\left( 1-e^{-x}\right)$
$\displaystyle \displaystyle \left( 1-e^{-x}\right)\sum_{n=1}^{\infty}\left( \frac{1}{e^{x}}\right)^n$
The series is geometric with $\displaystyle \displaystyle r= \frac{1}{e^{x}}$
So when does the geometric series converge uniformly.
No need for geometric series
Did you even bother with the hint given in the OP?
Use partial collapsing sums.
Let $\displaystyle \displaystyle S_N = \sum\limits_{n =1 }^N {\left[ {e^{ - nx} - e^{ - \left( {n + 1} \right)x} } \right] = e^{ - x} - e^{ - x\left( {N + 1} \right)} } $.
Now you have a rather simple sequence: $\displaystyle S_N = \left( {e^{ - x} - e^{ - x\left( {N + 1} \right)} } \right) \to e^{ - x} $
That is exactly the point!
If you let $\displaystyle x=0$ you get the series
$\displaystyle \displaystyle \sum_{n=1}^{\infty}[e^{-nx}-e^{-(n+1)x}]=\sum_{n=1}^{\infty}[1-1]=0$ so the series converges pointwise to zero there
Now by the geometric series when $\displaystyle x \ne 0$ the series converges to
$\displaystyle \displaystyle 1-e^{-x}\left( \frac{e^{-x}}{1-e^{-x}}\right)=e^{-x}$
so this sum converges to the function
$\displaystyle f(x)=\begin{cases}0 \, , \text{ if } x=0 \\ e^{-x}, \text{ if } x \in (0,1]\end{cases}$
Can a series of uniformly continuous functions converge to a discontinuous function?