Calculate limit of a sum

Question

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!

Answer 1

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 $\;k=1\;$ :

$$\sum_{k=1}^n\frac{(k-1)^7}{n^8}=\frac1n\sum_{k=0}^n\left(\frac kn\right)^7\xrightarrow[n\to\infty]{}\int\;\ldots$$

Answer 2

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:

$$\lim_{n\to+\infty}\sum_{k=0}^n \frac{\sqrt[n]{e^k}}{n}=\lim_{n\to+\infty}\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{1}{p+1}+\lim_{n\to+\infty}\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{R_p(n)}{n^{p+1}}=e-1+0=\color{blue}{e-1}.$$