Ok guys, it seems that I have become rusty since I cant find an apropriate way to tackle the following limit:

$\displaystyle \lim \limits_{n\to \infty}\frac{1}{n^2}\sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}$.

I really would appreciate any advice.

So far (I think) I have found both upper and lower boundary on the sum (probably not the best one) but thats all I got at the moment. Here goes...

$\displaystyle e^1\leq e^{1+\left(\frac{i}{n}\right)^2}\leq e^2, \qquad \forall n\in \mathbf{N}, i\in \mathbf{N},i\leq n.$

$\displaystyle \sum\limits_{i=1}^{n}i\cdot e\leq \sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq \sum\limits_{i=1}^{n}i\cdot e^2$

$\displaystyle e\cdot \sum\limits_{i=1}^{n}i \leq \sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq e^2\cdot \sum\limits_{i=1}^{n}i$

$\displaystyle e\cdot \frac{n(n+1)}{2} \leq \sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq e^2\cdot \frac{n(n+1)}{2}$

$\displaystyle \frac{e}{2}\cdot n^2 +\frac{e}{2}n\leq \sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq \frac{e^2}{2}\cdot n^2 +\frac{e^2}{2}n$

Now it seems one can conlude that

$\displaystyle \lim\limits_{n\to\infty}\frac{1}{n^2}\cdot \left( \frac{e}{2}\cdot n^2 +\frac{e}{2}n\right) \leq \lim \limits_{n\to \infty}\frac{1}{n^2}\sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq \lim\limits_{n\to \infty} \frac{1}{n^2}\cdot\left(\frac{e^2}{2}\cdot n^2 +\frac{e^2}{2}n\right) $.

That means that $\displaystyle \frac{e}{2}\leq \lim \limits_{n\to \infty}\frac{1}{n^2}\sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq \frac{e^2}{2}$.

So even if I'm right with all the above (which is not necessarily the case ) , that still doesn't give me the exact value of the limit, so I'm pretty much stuck. I hope it is something trivial that I'm overlooking and that someone can direct me, give me a push, towards the solution.

Thanks in advance, MathoMan.

PS. I have tried using Wolfram Mathematica and Wolfram Alpha but they both went numb on this one. So I evaluated the expression under the limit operator for n=100000 and got the number 2.335424 and also for n=1000000 in which case the outcome was 2.33549083. So I'm pretty much out of ideas.