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Find the limit: $\lim_{n \to +\infty}(\frac{1^p + 2^p + ... + n^p}{n^p} - \frac{n}{p + 1})$, where $p \in \mathbb{N}$.

I've got an idea to transform this sequence into $lim_{n\to +\infty}(\frac{x_n}{y_n})$ and to use Stolz's theorem.

But in 2 days I didn't get ideas for this transformation.

Someone
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  • Do you know about Faulhaber's formula? The limit turns out to be $1/2$ since the remainder terms have a pth degree polynomial in n in the denominator which diverges to infinity and thus the remainder terms vanish asymptotically. – Prasun Biswas Nov 04 '20 at 20:32
  • Have you at least worked out the cases $p = 1$, $2$, $3$ (which you have not shown in your Question) so that you know what answer you are supposed to get? – Eric Towers Nov 04 '20 at 20:36
  • Yes, I tried. And in these terms limit equals to $\frac{1}{2}$. But that didn't help me come to the correct reasoning. – Someone Nov 04 '20 at 20:52

1 Answers1

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Hint:

You should try showing that, in general:

$$\sum_{k=1}^{n} k^p = \frac{n^{p+1}}{p+1} + \frac{1}{2}n^p+ q(n)$$

where $q(n)$ is a polynomial in $n$ of the $(p-1)$-th degree. Can you use this to simplify the expression that you're taking the limit of?

Abhijeet Vats
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  • Oh, looks too hard for me in terms what I have to know right now. I will try to figure it out, but perhaps this hint does not help me to come to a solution. – Someone Nov 04 '20 at 20:48