# Alternative proof of the limitof the quotient of two sums.

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

$$\lambda_n(a)=\frac{s_n(a)}{n^{a+1}}$$

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.)

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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

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 $$\lim_{n\to +\infty} \frac{s_n(a+1)}{ns_n(a)}=\left\{\begin{array}{cl}\frac{a+1}{a+2}&\text{when }a>-1\\0&\text{when }a\le-1\end{array}\right.\tag{7}$$

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

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