# Function Series pointwise/uniform/compact convergence

In the following excercise I have to tell whether the function-series is a) pointwise convergent b) uniform convergent and/or c) compact convergent on intervall $[a, \infty]$ (a > 0).

The series is:

$$\sum_{n=1}^\infty x \cdot [ n^2\cdot e^{-(n+1)x} - (n-1)^2\cdot e^{-nx}].$$

The problem is: I really don't know what I have to do. So I'll be glad for any kind of help :)

a) The series is pointwise convergent, if the function-sequence is pointwise convergent. If my series is pointwise convergent and therefore the sequence as well, I should be able to find the "pointwise limit of the sequence" and this means I have to find a function $f = \lim f_n$. Is this correct so far?

b) The series is uniform convergent, if the function-sequence is uniform convergent. In this case, the sequence $(f_n)$ converges to a function $f$, if $\forall \varepsilon > 0 \exists n_0 \in \mathbb{N} \forall n \ge n_0 \forall x \in M: |f_n(x) - f(x)| < \varepsilon$.?

c) ?

Define $a_n(x):=n^2\exp(-(n+1)x)$ and notice that we have to study the convergence of $\sum_n a_n(x)-a_{n-1}(x)$, hence we have an explicit formula for partial sums.
There is pointwise convergence on $[0,\infty)$ but not uniform (take $x_n:=1/n$).
• Hello! Well $\sum_ a_n(x)−a_{n−1}(x) = a_n(x)$ right? (Telescope sum). Therefore I use the ratio test; $lim| \frac{a_{n+1}}{a_n}| = e^{-x}\cdot \frac{(n+1)^2}{n^2}$. There should be something wrong. Should I use a different test? – Vazrael Jan 13 '14 at 9:21