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What is $$\lim\limits_{n\to\infty}\sqrt[n]{1^k +2^k+\cdots+ n^k},$$ where $ k\in \mathbb{N}$?

jimjim
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Julia
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    Are you asking about limits in general, or that one limit in particular? – Arthur Nov 16 '20 at 22:35
  • You might want to check out Faulhaber's formula (https://en.wikipedia.org/wiki/Faulhaber%27s_formula), which gives a formula for the expression under the $n$th root. – aras Nov 16 '20 at 22:38
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2 Answers2

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Hint: $$1\le 1^k +2^k+\cdots+ n^k\le n^{k+1}$$ and $$\lim_{n\to\infty}n^{1/n}=1.$$

Bumblebee
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When we deal with limits involving the $n^{\text{th}}$ root of an expression, a useful result is this one: https://math.stackexchange.com/a/3159844/629594.
With this in mind, we may write that $\displaystyle\lim_{n\to\infty}\sqrt[n]{1^k +2^k+\cdots+ n^k}=\lim_{n\to \infty}\frac{1^k+2^k+...+(n+1)^k}{1^k+2^k+...+n^k}\stackrel{\text{Stolz-Cesaro}}{=}\lim_{n\to \infty} \frac{(n+2)^k}{(n+1)^k}=1$.
The other answer is nicer, because it just uses the squeeze theorem, but I wanted to show that this can roughly be done by applying Stolz-Cesaro two times.

Alexdanut
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