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On a limit with summation

Let S_k(n) = \sum \limits_{m=1}^{n} m^k \; , \; k \in \mathbb{N}^*. Prove that

    \[\lim_{n \rightarrow +\infty} n \left ( \frac{S_k(n)}{n^{k+1}} - \frac{1}{k+1} \right ) = \frac{1}{2}\]


Let us begin with the simple observation that

    \begin{align*} \lim_{n \rightarrow +\infty} \frac{1}{n^{k+1}} \sum_{m=1}^{n} m^k &=\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{m=1}^{n} \left ( \frac{m}{n} \right )^k \\ &=\int_{0}^{1} x^k \, \mathrm{d}x \\ &= \frac{1}{k+1} \end{align*}

Now, here comes a handy lemma.

Lemma: Let f:[0, 1] \rightarrow \mathbb{R} be a differentiable function with continuous derivative. It holds that

    \[\lim_{n\rightarrow +\infty}n \left(\frac{1}{n}\sum_{i=1}^{n}f \left(\frac{i}{n} \right)-\int_{0}^{1}f(x) \, \mathrm{d}x \right)= \frac{f(1)-f(0)}{2}\]

Proof: The derivation of the theorem follows from application of MVT in the interval \left[x , \frac{i}{n} \right].

Hence the limit follows to be \frac{1}{2}.

Note: Applying Euler  – MacLaurin we have

    \[\sum_{m=0}^{n} m^k = \frac{n^{k+1}}{k+1} + \frac{n^k}{2} + \frac{k n^{k-1}}{12} + \mathcal{O}\left(n^{k-3}\right)\]

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