Find the value of $ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left( \frac{k}{n^2}\right )^{\frac{k}{n^2} +1} $ Compute limit of sum :
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left( \frac{k}{n^2} \right)^{\dfrac{k}{n^2} +1} $$
My Attempt :
$$\Big( \dfrac{k}{n^2} \Big)^{\frac{k}{n^2} +1} = e^{\Big( \frac{k}{n^2} +1 \Big) \log\Big(\frac{k}{n^2}\Big)} = \Big( e^{\Big( \frac{k}{n^2} +1 \Big) }\Big)^{\log\Big(\frac{k}{n^2}\Big)} = \dfrac{1}{n} \cdot \dfrac{k}{n} \Big( e^{\frac{k}{n^2}} \Big)^{\log (\frac{k}{n^2})}  $$
I am unable to move ahead, please help.
 A: Trying to solve my problem moving along with Winther's idea (thanks a lot) :
$\Big( \dfrac{k}{n^2} \Big)^{\frac{k}{n^2} +1} = e^{\Big( \frac{k}{n^2} +1 \Big) \log\Big(\frac{k}{n^2}\Big)} = \Big( e^{\Big( \frac{k}{n^2} +1 \Big) }\Big)^{\log\Big(\frac{k}{n^2}\Big)} = \dfrac{1}{n} \cdot \dfrac{k}{n} \Big( e^{\frac{k}{n^2}} \Big)^{\log (\frac{k}{n^2})} $
$= \dfrac{1}{n} \cdot \dfrac{k}{n} \Bigg(1 + \mathcal{O} \Big( \frac{k}{n^2}\log (\frac{k}{n^2}) \Big) \Bigg)  $
we want to use the given series expansion for small $x$ :
$x^x \sim \sum_{k = 0}^{\infty} \frac{1}{k!} (x \log x)^k = 1 + x \log x + \frac{1}{2} x^2 \log^2 x + O(x^3 \log^3 x) .$
first term obviously gives :
$\displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \dfrac{1}{n} \cdot \dfrac{k}{n} = \int_0^1 xdx = \dfrac{1}{2}$
for second term we use :
$ \dfrac{k}{n^2} \mathcal{O} \Big( \frac{k}{n^2}\log (\dfrac{k}{n^2}) \Big) =  \Big[ \Big(\dfrac{k}{n^2}\Big)^2 \log \Big( \dfrac{k}{n^2} \Big) + ... \Big]$
where, $ n \cdot \Big(\dfrac{1}{n^2}\Big)^2 \log \Big( \dfrac{1}{n^2} \Big) < \displaystyle\sum_{k=1}^n \Big(\dfrac{k}{n^2}\Big)^2 \log \Big( \dfrac{k}{n^2} \Big) < n \cdot \Big(\dfrac{n}{n^2}\Big)^2 \log \Big( \dfrac{n}{n^2} \Big)$
using squeez theorem, we see, both sides sum will go to $0$ as $n \rightarrow \infty$ and similarly the higher order terms as well.
hence, answer is = $\dfrac{1}{2}$.
