Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I found the following problem by Apostol: Let $a \in \Bbb R$ and $s_n(a)=\sum\limits_{k=1}^n k^a$. Find

$$\lim_{n\to +\infty} \frac{s_n(a+1)}{ns_n(a)}$$

After some struggling and helpless ideas I considered the following solution.

If $a > -1$, then

$$\int_0^1 x^a dx=\frac{1}{a+1}$$ is well defined. Thus, let


It is clear that

$$\lim\limits_{n\to +\infty} \lambda_n(a)=\int_0^1 x^a dx=\frac{1}{a+1}$$

and thus

$$\lim_{n\to +\infty} \frac{s_n(a+1)}{ns_n(a)}=\lim_{n \to +\infty} \frac{\lambda_n(a+1)}{\lambda_n(a)}=\frac{a+1}{a+2}$$

Can you provide any other proof for this? I used mostly integration theory but maybe there are other simpler ideas (or more complex ones) that can be used.

(If $a=-1$ then the limit is zero, since it is simply $H_n^{-1}$ which goes to zero since the harmonic series is divergent. For the case $a <-1$, the simple inequalities $s_n(a+1) \le n\cdot n^{a+1} = n^{a+2}$ and $s_n(a) \ge 1$ show that the limit is also zero.)

share|cite|improve this question
You can use standard results from the theory of finite differences. The partial summation of a polynomial of degree $d$ with leading term $c$ is a polynomial of degree $d+1$ with leading term $\frac{c}{d+1}$. – Qiaochu Yuan May 26 '12 at 15:14
@QiaochuYuan Does that work for any real $a$? – Pedro Tamaroff May 26 '12 at 16:35
What about Stolz-Cesaro theorem? (Although it would give a very similar proof.) – Martin Sleziak May 26 '12 at 21:23
@MartinSleziak Go ahead! =) I tried, but then I realized I could solve it this way so I left it inconclusive. – Pedro Tamaroff May 26 '12 at 22:07
I meant using Stolz-Cesaro to shot $\lim_{n\to\infty}\frac{s_n(a)}{n^{a+1}}=\lim_{n\to\infty}\frac{\sum k^a}{n^{a+1}}=\lim_{n\to\infty}\frac{n^a}{n^{a+1}-(n-1)^{a+1}}=\frac1{a+1}$ and rewrite the original fraction as $\frac{s_n(a+1)}{n^{a+2}}\frac{n^a}{s_n(a)}$. Which is basically the same as your solution, only in the place where you used integral, I used Stolz Cesaro. (If we wanted to use Stolz-Cesaro for the limit in the given form, we would probably have to use it twice and it would be more complicated.) – Martin Sleziak May 27 '12 at 3:35
up vote 2 down vote accepted

For $a\ge0$, $x^a$ is a monotonic non-decreasing function, therefore, $$ \small\frac{1}{a+1}\left[n^{a+1}-0\right]=\int_0^nx^a\,\mathrm{d}x\le\sum_{k=1}^nk^a\le\int_1^{n+1}x^a\,\mathrm{d}x=\frac{1}{a+1}\left[(n+1)^{a+1}-1\right]\tag{1} $$ For $-1< a<0$, $x^a$ is a monotonic non-increasing function, therefore, $$ \small\frac{1}{a+1}\left[n^{a+1}-0\right]=\int_0^nx^a\,\mathrm{d}x\ge\sum_{k=1}^nk^a\ge\int_1^{n+1}x^a\,\mathrm{d}x=\frac{1}{a+1}\left[(n+1)^{a+1}-1\right]\tag{2} $$ Combining $(1)$ and $(2)$ yields that for $a>-1$ $$ \lim_{n\to\infty}\frac{1}{n^{a+1}}s_n(a)=\frac{1}{a+1}\tag{3} $$ $x^{-1}$ is a monotonic non-increasing function, therefore, $$ 1+\log(n)=1+\int_1^nx^{-1}\,\mathrm{d}x\ge\sum_{k=1}^nk^{-1}\ge\int_1^{n+1}x^{-1}\,\mathrm{d}x=\log(n+1)\tag{4} $$ Thus, $$ \lim_{n\to\infty}\frac{1}{\log(n)}s_n(-1)=1\tag{5} $$ For $a<-1$, $$ \lim_{n\to\infty}s_n(a)=\zeta(-a)\tag{6} $$ Combining $(3)$, $(5)$, and $(6)$ yields $$ \lim_{n\to +\infty} \frac{s_n(a+1)}{ns_n(a)}=\left\{\begin{array}{cl}\frac{a+1}{a+2}&\text{when }a>-1\\0&\text{when }a\le-1\end{array}\right.\tag{7} $$

share|cite|improve this answer

The argument below works for any real $a > -1$. We are given that $$s_n(a) = \sum_{k=1}^{n} k^a$$ Let $a_n = 1$ and $A(t) = \displaystyle \sum_{k \leq t} a_n = \left \lfloor t \right \rfloor$. Hence, $$s_n(a) = \int_{1^-}^{n^+} t^a dA(t)$$ The integral is to be interpreted as the Riemann Stieltjes integral. Now integrating by parts, we get that $$s_n(a) = \left. t^a A(t) \right \rvert_{1^-}^{n^+} - \int_{1^-}^{n^+} A(t) a t^{a-1} dt = n^a \times n - a \int_{1^-}^{n^+} \left \lfloor t \right \rfloor t^{a-1} dt\\ = n^{a+1} - a \int_{1^-}^{n^+} (t -\left \{ t \right \}) t^{a-1} dt = n^{a+1} - a \int_{1^-}^{n^+} t^a dt + a \int_{1^-}^{n^+}\left \{ t \right \} t^{a-1} dt\\ = n^{a+1} - a \left. \dfrac{t^{a+1}}{a+1} \right \rvert_{1^-}^{n^+} + a \int_{1^-}^{n^+}\left \{ t \right \} t^{a-1} dt\\ =n^{a+1} - a \dfrac{n^{a+1}-1}{a+1} + a \int_{1^-}^{n^+}\left \{ t \right \} t^{a-1} dt\\ = \dfrac{n^{a+1}}{a+1} + \dfrac{a}{a+1} + \mathcal{O} \left( a \times 1 \times \dfrac{n^a}{a}\right)\\ = \dfrac{n^{a+1}}{a+1} + \mathcal{O} \left( n^a \right)$$ Hence, we get that $$\lim_{n \rightarrow \infty} \dfrac{s_n(a)}{n^{a+1}/(a+1)} = 1$$ Hence, now $$\dfrac{s_{n}(a+1)}{n s_n(a)} = \dfrac{\dfrac{s_n(a+1)}{n^{a+2}/(a+2)}}{\dfrac{s_n(a)}{n^{a+1}/(a+1)}} \times \dfrac{a+1}{a+2}$$ Hence, we get that $$\lim_{n \rightarrow \infty} \dfrac{s_{n}(a+1)}{n s_n(a)} = \dfrac{\displaystyle \lim_{n \rightarrow \infty} \dfrac{s_n(a+1)}{n^{a+2}/(a+2)}}{\displaystyle \lim_{n \rightarrow \infty} \dfrac{s_n(a)}{n^{a+1}/(a+1)}} \times \dfrac{a+1}{a+2} = \dfrac11 \times \dfrac{a+1}{a+2} = \dfrac{a+1}{a+2}$$

Note that the argument needs to be slightly modified for $a = -1$ or $a = -2$. However, the two cases can be argued directly itself.

If $a=-1$, then we want $$\lim_{n \rightarrow \infty} \dfrac{s_n(0)}{n s_n(-1)} = \lim_{n \rightarrow \infty} \dfrac{n}{n H_n} = 0$$

If $a=-2$, then we want $$\lim_{n \rightarrow \infty} \dfrac{s_n(-1)}{n s_n(-2)} = \dfrac{6}{\pi^2} \lim_{n \rightarrow \infty} \dfrac{H_n}{n} = 0$$

In general, for $a <-2$, note that both $s_n(a+1)$ and $s_n(a)$ converge. Hence, the limit is $0$. For $a \in (-2,-1)$, $s_n(a)$ converges but $s_n(a+1)$ diverges slower than $n$. Hence, the limit is again $0$.

Hence to summarize $$\lim_{n \rightarrow \infty} \dfrac{s_n(a+1)}{n s_n(a)} = \begin{cases} \dfrac{a+1}{a+2} & \text{ if }a>-1\\ 0 & \text{ if } a \leq -1 \end{cases}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.