Evaluate $ \lim_{n\to+\infty}\sum_{k=1}^{n}(e^{\frac{k^2}{n^3}}-1)$. 
Evaluate
  $$
\lim_{n\to+\infty}\sum_{k=1}^{n}(e^{\frac{k^2}{n^3}}-1).
$$

Since $\sum_{k=1}^{n}(e^{\frac{k^2}{n^3}}-1)\leq n(e^{\frac{1}{n}})-n$, which implies that the limit is no more than $1$. But I met some problems in the method of enlarging and reducing.
 A: As you refer to reducing and enlarging, squeezing is a clean way here together with Riemann sums:
To do so, apply Taylor to $e^x$ with 2nd degree remainder and bound the remainder:


*

*$x> 0 \Rightarrow e^x-1 = x+ \frac{e^{\xi}}{2}x^2 \mbox{ with } 0<\xi <x$

*$\stackrel{0 \leq x\leq 1}{\Rightarrow} x + \frac{1}{2}x^2 \leq e^x-1 \leq x + \frac{e}{2}x^2$
Applying this to your sum you get
$$\underbrace{\sum_{k=1}^n\left(\frac{k}{n}\right)^2\frac{1}{n}}_{\stackrel{n\to \infty}{\longrightarrow}\int_0^1x^2\;dx} + \frac{1}{2n}\underbrace{\sum_{k=1}^n\left(\frac{k}{n}\right)^4\frac{1}{n}}_{\stackrel{n\to \infty}{\longrightarrow}\int_0^1x^4\;dx} \leq \sum_{k=1}^n\left(e^{\frac{k^2}{n^3}}-1\right) \leq \underbrace{\sum_{k=1}^n\left(\frac{k}{n}\right)^2\frac{1}{n}}_{\stackrel{n\to \infty}{\longrightarrow}\int_0^1x^2\;dx} + \frac{e}{2n}\underbrace{\sum_{k=1}^n\left(\frac{k}{n}\right)^4\frac{1}{n}}_{\stackrel{n\to \infty}{\longrightarrow}\int_0^1x^4\;dx}$$
Now, sending $n\to \infty$ gives $\lim_{n\to \infty}\sum_{k=1}^n\left(e^{\frac{k^2}{n^3}}-1\right) = \int_0^1x^2\;dx = \frac{1}{3}$.
A: We have that
$$e^{\frac{k^2}{n^3}}-1=\frac{k^2}{n^3}+O\left(\frac{k^4}{n^6}\right)$$
and therefore
$$\sum_{k=1}^{n}(e^{\frac{k^2}{n^3}}-1)=\frac1{n^3}\sum_{k=1}^{n}k^2+\frac1{n^6}\sum_{k=1}^{n}O(k^4)$$
then refer to Faulhaber's formula
A: $$L=\lim_{n \rightarrow \infty}\sum_{k=1}^{n} (e^{k^2/n^3}-1)= \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \left( [(k^2/n^2)+ \frac{1}{2}(k^4/n^5)+... ]\right).$$ 
Let $k/n=x$, then  $k^4/n^5 \rightarrow 0$, when $n\rightarrow \infty.$
Then $$L=\int_{0}^{1} x^2 dx=\frac{1}{3}.$$
