Calculate limit of a sum I'm currently repeating for exam and i'm stuck with limits of following two sums.
$$\lim_{n\rightarrow +\infty} \sum_{k=0}^n \frac{(k-1)^7}{n^8}$$
and
$$\lim_{n\rightarrow +\infty} \sum_{k=0}^n \frac{\sqrt[n]{e^k}}{n}$$
Maybe if sum were from $k=0$ to $\infty$ then i could change it to integral and then calculate it somehow, but i this is a first time i see such task and i'd be greatful for ideas how to solve such tasks... 
Thank you in advance!
 A: Hints:
$$\sum_{k=0}^n\frac{\sqrt[n]{e^k}}n=\frac1n\sum_{k=0}^ne^{k/n}\xrightarrow[n\to\infty]{}\int\limits_0^1e^x\,dx$$
I'm assuming the first sum actually begins at $\;k=1\;$ :
$$\sum_{k=1}^n\frac{(k-1)^7}{n^8}=\frac1n\sum_{k=0}^n\left(\frac kn\right)^7\xrightarrow[n\to\infty]{}\int\;\ldots$$
A: I did it this way:
Using Faulhaber's formula we have:
$$\sum_{k=1}^n k^p = {1 \over p+1} \sum_{j=0}^p (-1)^j{p+1 \choose j} B_j n^{p+1-j},\qquad \mbox{where}~B_1 = -\frac{1}{2}.$$
From this, we can see the $\sum\limits_{k=1}^{n}k^p$ is a polynomial in $n$ with degree $p+1$ and corresponding coefficient $\dfrac{1}{p+1}$.
Thus,
$$\sum_{k=0}^n \frac{(k-1)^7}{n^8}=\sum_{k=0}^n \frac{k^7}{n^8}+ \frac{P(n)}{n^8},$$
where $P(n)$ is a polynomial in $n$ with degree $\leq 7$. Further:
$$\sum_{k=0}^n \frac{k^7}{n^8}+ \frac{P(n)}{n^8}= \frac{n^8}{(1+7)n^8}+ \frac{Q(n)}{n^8}+\frac{P(n)}{n^8},$$
where $Q(n)$ is a polynomial in $n$ with degree $\leq7$.
Therefore:
$$\lim_{n\to+\infty}\sum_{k=0}^n \frac{(k-1)^7}{n^8}=\frac{1}{(1+7)}=\color{blue}{\frac{1}{8}}.$$
The same strategy is applied for the next summation.
$$\sum_{k=0}^n \frac{\sqrt[n]{e^k}}{n}=\sum_{k=0}^n \frac{{e^{k/n}}}{n}=\sum_{k=0}^n\sum_{p=0}^\infty \dfrac{1}{n}\frac{{(k/n)^p}}{p!}=\sum_{p=0}^\infty\dfrac{1}{p!n^{p+1}}\sum_{k=0}^nk^p.$$
Thus,
$$\sum_{k=0}^n \frac{\sqrt[n]{e^k}}{n}=\sum_{p=0}^\infty\dfrac{1}{p!n^{p+1}}\dfrac{n^{p+1}}{p+1}+\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{R_p(n)}{n^{p+1}}=\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{1}{p+1}+\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{R_p(n)}{n^{p+1}}=\sum_{p=0}^\infty\dfrac{1}{(p+1)!}+\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{R_p(n)}{n^{p+1}},$$
where $R_p(n)$ is a polynomial in $n$ with degree $\leq p$.
Therefore:
$$\lim_{n\to+\infty}\sum_{k=0}^n \frac{\sqrt[n]{e^k}}{n}=\lim_{n\to+\infty}\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{1}{p+1}+\lim_{n\to+\infty}\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{R_p(n)}{n^{p+1}}=e-1+0=\color{blue}{e-1}.$$
