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I know Stolz-Cesaro theorem, and I'm supposed to use this to prove that

$\lim(1^p+2^p+...+n^p)/(n^{p+1})=1/(p+1)$

So I made two sequences, $(x_n)$ which is:

$\sum_{i=1}^n i^p$

And $(y_n)$ which is $n^{p+1}$

So I'm trying to prove $\lim(x_{n+1}- x_n)/ (y_{n+1}-y_n)$

And after some work I get to

$1/(1-((n/n+1)^p)$,

But I don't know what to do when here. It's possible that this approach isn't even right. $n$ tends to infinity. Thanks in advance

Martin Sleziak
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Jeff
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  • See also: http://math.stackexchange.com/questions/419765/limit-lim-n-to-infty-frac1p2p-ldotsnpnp1, http://math.stackexchange.com/questions/589274/limit-of-a-riemann-sum-lim-n-to-infty-frac1p2p-ldotsnpnp1, http://math.stackexchange.com/questions/478344/what-is-the-result-of-lim-n-to-infty-frac-sumn-i-1-iknk1 and other posts linked there. – Martin Sleziak Oct 14 '14 at 06:32

1 Answers1

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Hint After SC you get

$$\frac{(n+1)^p}{(n+1)^{p+1}-n^{p+1}}$$

You probably tried to divide by $(n+1)^p$ but that is unnecessary and makes the problem harder. Just expand $(n+1)^{p+1}=n^{p+1}+(p+1)n^p+\mbox{junk}$ in the denominator and cancel $n^{p+1}-n^{p+1}$

N. S.
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  • Why am I allowed to use that? – Jeff Oct 13 '14 at 19:49
  • @George What do you mean? What do you think you are not allowed to do and why? – N. S. Oct 13 '14 at 19:50
  • The expansion on the denominator. I'm sure that it is true, I've just never seen it before so I wonder where it comes from is what I mean – Jeff Oct 13 '14 at 19:52
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    @George http://en.wikipedia.org/wiki/Binomial_theorem – N. S. Oct 13 '14 at 19:53
  • @George Or just multiply $(n+1)(n+1)...(n+1)$ by expanding. Note that what you get if you expand is a $n^{p+1}$, when you chose $n$ from each bracket, some $n^p$ when you chose $p$ n's and one one, and smaller degree terms. How many $n^p$ do you get? – N. S. Oct 13 '14 at 19:55
  • @N.S. What can I do with the remaining junk then? – Britanica Jan 01 '21 at 10:42