A familiar limit, but in general form .

Question

Suppose $a \in \mathbb{R} ,a>1$ Is there an idea to compute the limit below ? $$\lim_{n \to \infty}\left( \frac{1^a+2^a+3^a+...+n^a}{n^a}-\frac{n}{a+1} \right)$$ I tried it for $a=1,2,3$ but I get stuck in general form . Can someone help me ? Thanks in advance.

Answer 1

By Stolz theorem (see here: https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) we obtain: $$\lim_{n \to \infty}\left( \frac{1^a+2^a+3^a+...+n^a}{n^a}-\frac{n}{a+1} \right)=$$ $$=\frac{1}{a+1}\lim_{n\rightarrow+\infty}\frac{(a+1)(1^a+2^a+3^a+...+n^a)-n^{a+1}}{n^a}=$$ $$=\frac{1}{a+1}\lim_{n\rightarrow+\infty}\frac{(a+1)n^a-n^{a+1}+(n-1)^{a+1}}{n^a-(n-1)^a}=$$ $$=\frac{1}{a+1}\lim_{n\rightarrow+\infty}\frac{(a+1)n^a-n^{a+1}+n^{a+1}-(a+1)n^a+\frac{a(a+1)}{2}n^{a-1}+...}{n^a-n^a+an^{a-1}+...}=\frac{1}{2}.$$

Answer 2

You may also apply basic Combinatorics. $n^a$ is the number of functions from $[1,a]$ to $[1,n]$ and $k!{n\brace k}$ (where the Stirling number of the second kind ${n\brace k}$ accounts for the number of ways of partitioning $[1,n]$ into $k$ non-empty subsets) is the number of surjective functions from $[1,n]$ to $[1,k]$. By classifying the functions from $[1,a]$ to $[1,n]$ according to the cardinality of their range we have $$ n^a = \sum_{h=1}^{a}\binom{n}{h}h!{a\brace h} $$ and by the hockey-stick identity $$\begin{eqnarray*} \sum_{n=1}^{N} n^a &=& \sum_{h=1}^{a}\binom{N+1}{h+1}h!{a\brace h}\\&=&\binom{N+1}{a+1}a!+\binom{N+1}{a}(a-1)!\frac{a(a-1)}{2}+\sum_{h<a-1}\binom{N+1}{h+1}h!{a\brace h}. \end{eqnarray*}$$ It follows that $$\begin{eqnarray*} \frac{1}{N^a}\sum_{n=1}^{N} n^a&=&\frac{(N+1)\cdots(N-a+1)}{N^a(a+1)}+\frac{(N+1)\cdots(N-a+2)\cdot(a-1)}{2\cdot N^a}+O\left(\frac{1}{N}\right) \end{eqnarray*}$$ and simplifying $$\begin{eqnarray*} \frac{1}{N^a}\sum_{n=1}^{N} n^a &=& \frac{N+1}{a+1}\prod_{k=1}^{a-1}\left(1-\frac{k}{N}\right)+\frac{a-1}{2}+O\left(\frac{1}{N}\right)\\&=&\frac{N+1}{a+1}\left(1-\frac{a(a-1)}{2N}\right)+\frac{a-1}{2}+O\left(\frac{1}{N}\right)\\&=&\frac{N}{a+1}+\color{red}{\frac{1}{2}}+O\left(\frac{1}{N}\right).\end{eqnarray*}$$ Indeed, the whole problem boils down to the computation of the Bernoulli number $B_1$, as already remarked in the comment about the EML formula.