the partial sum of a series of functions? For the function $f_n(x) = x e^{-nx^2}\bigl((n+1) e^{-x^2}-n\bigr),~~ \forall n \geq 0$ and $\forall x \in [0,1]$.
We define the partial sum as $N_n(x) = \sum_{i=0}^n f_i(x)$
I'm trying to find the partial sum in order to compare 
$$ \lim_{n \rightarrow \infty } \int_0^1 N_n(x) dx~~ \text{and}  \int_0^1 \lim_{n \rightarrow \infty } N_n(x) dx $$
The problem is that I can't simplify and find the partial sum. I tried to simplify it as 
\begin{eqnarray}
N_n(x) = \sum_{i=0}^n f_i(x) & = & \sum_{i=0}^n x e^{-ix^2}\bigl((i+1) e^{-x^2}-i\bigr) \nonumber  \\
& = & \sum_{i=0}^n \bigl[ x(i+1) e^{-x^2(1+i)} - ixe^{-ix^2}\bigr] \nonumber \\
& = & x e^{-x^2}\sum_{i=0}^n  (i+1) e^{-ix^2} - x \sum_{i=0}^n ie^{-ix^2} \nonumber \\
\end{eqnarray}
How would I continue to find the partial sum? Also, what is the idea of the question? i.e what is the idea of comparing th limit of integral to the integral of the limit ?
Update:
using the telescoping from the hint in the answer below, the sum equals $(n+1) x e^{-(n+1)x^2}$. 
For the other part of my question regarding the comparison, I found $$ \lim_{n \rightarrow \infty } \int_0^1 N_n(x) dx =  \frac{1}{2}\lim_{n \rightarrow \infty } \bigl(1- e^{-(n+1)} \bigr) = \frac{1}{2}$$ and the other one is $$ \int_0^1 \bigl( \lim_{n \rightarrow \infty }  N_n(x) \bigr) dx =  \int_0^1 \bigl( \lim_{n \rightarrow \infty }  (n+1) x e^{-(n+1)x^2} \bigr)dx  = \int_0^1 0dx = 0 $$
SO WHAT? What is the idea of a similar comparison generally ? 
 A: In the last but one line of your derivation you have a telescoping sum: so you can conclude easily.   
-- addendum --
In reply to your comment :
 - yes then the sum equals $
N_{\,n}  = \left( {n + 1} \right)xe^{\, - \left( {n + 1} \right)x^{\,2} } 
$;
 - in the expression with the  limit inside the integral you are going to have a "$\infty \cdot 0$ situation; if you plot the integrand function you can see what's going to happen

It tells you that the function $N_n(x)$ is not going  flatly to zero, but peaks to infinity near $0^+  \leftarrow x$ before decaying rapidly to zero.
And it does so while keeping a finite integral of $1/2$ as you discovered with the "external" limit.
In fact the function has a maximum at
$$
\eqalign{
  & 0 = {d \over {dx}}N_{\,n} (x) = \left( {1 - 2\left( {n + 1} \right)x^{\,2} } \right)\left( {n + 1} \right)e^{\, - \left( {n + 1} \right)x^{\,2} } \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \max :\quad x_{\,M}  = {1 \over {\sqrt {2\left( {n + 1} \right)} }}\quad N_{\,n} (x_{\,M} ) = \sqrt {{{\left( {n + 1} \right)} \over 2}} e^{\, - 1/2}  \cr} 
$$
In conclusion, for $n \to \infty$, $N_n(x)$ becomes proportional to the derivative of the Dirac Delta function, in its representation 
as the limit of a Normal distribution.
