sequences and series - Calculate the limit: $limlimits_{ntoinfty} a_n = frac{1}{n^2}sumlimits_{i=1}^nicdot e^frac{i^2}{n^2}$

Calculate the limit of the following sequence: $\lim\limits_{n\to \infty} a_n = \frac{1}{n^2}\sum\limits_{i=1}^ni\cdot e^\frac{i^2}{n^2}$

Any help or hint would be appreciated.

Answer

Write
$$a_n := \frac{1}{n^2} \sum \limits_{k = 1}^n k \cdot e^{\frac{k^2}{n^2}} = \frac{1}{n} \sum \limits_{k = 1}^n \frac{k}{n} \cdot e^{(\frac{k}{n})^2}$$
This is a Riemann sum for the (everywhere defined and continuous!) function $f(x) := x \cdot e^{x^2}$ over the interval $[0,1]$, hence the limit you're looking for exists and is given by
$$\lim \limits_{n \to \infty} a_n = \int \limits_0^1 x \cdot e^{x^2} dx$$
Elementary integration (e.g. using the substitution $y = x^2$) yields
$$\lim \limits_{n \to \infty} a_n = \int \limits_0^1 x \cdot e^{x^2} dx = \left[ \frac{1}{2} e^{x^2} \right]_0^1 = \frac{1}{2}(e-1)$$