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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!

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This formula may help you. – x.y.z... May 7 '14 at 18:51

2 Answers 2

up vote 5 down vote accepted



I'm assuming the first sum actually begins at $\;k=1\;$ :

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

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Oh man, such a beautiful move! Thank you for this – Spinach May 7 '14 at 19:03
if i have $$\sum_{k=1}^n ln((n+k)^{n+k}) \xrightarrow[n\to\infty]{} \int_{ln 4}^\infty x dx$$ is that right? According to that, this sum diverges, right? – Spinach May 7 '14 at 19:14
First, I can't see here how that sum equals that integral, second: it's easy to see $\;\log(n+k)^{n+k}\rlap{\;\;\;/}\longrightarrow 0\;$ and thus the series can't converge... – DonAntonio May 7 '14 at 19:17
Yes yes you are right... I wasn't right cause $n,k$ aren't real numbers. Thanks again – Spinach May 7 '14 at 19:39

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 $\dfrac{1}{p+1}$.


$$\sum_{k=0}^n \frac{(k-1)^7}{n^8}=\sum_{k=0}^n \frac{k^7}{n^8}+ \frac{P(n)}{n^8},$$ 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 $\leq7$.


$$\lim_{n\to+\infty}\sum_{k=0}^n \frac{(k-1)^7}{n^8}=\frac{1}{(1+7)}=\color{blue}{\frac{1}{8}}.$$

The same strategy is applied for the next summation.

$$\sum_{k=0}^n \frac{\sqrt[n]{e^k}}{n}=\sum_{k=0}^n \frac{{e^{k/n}}}{n}=\sum_{k=0}^n\sum_{p=0}^\infty \dfrac{1}{n}\frac{{(k/n)^p}}{p!}=\sum_{p=0}^\infty\dfrac{1}{p!n^{p+1}}\sum_{k=0}^nk^p.$$


$$\sum_{k=0}^n \frac{\sqrt[n]{e^k}}{n}=\sum_{p=0}^\infty\dfrac{1}{p!n^{p+1}}\dfrac{n^{p+1}}{p+1}+\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{R_p(n)}{n^{p+1}}=\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{1}{p+1}+\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{R_p(n)}{n^{p+1}}=\sum_{p=0}^\infty\dfrac{1}{(p+1)!}+\sum_{p=0}^\infty\dfrac{1}{p!}\dfrac{R_p(n)}{n^{p+1}},$$

where $R_p(n)$ is a polynomial in $n$ with degree $\leq p$.


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

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