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I need to find the limit of a sequence indexed by $n\in \Bbb N$. $k$ is fixed natural constant.

The sequence is:

$x_n = \frac{1^k+2^k+3^k+\ldots+n^k}{n^k} - \frac{n}{k+1}$

I tried to solve this problem with Stolz–Cesàro theorem. But I can't find the limit.

While trying to code this equation on Python I've found the limit is about -0.5 from top $(-0.5+\epsilon)$

Gabriel Romon
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franchb
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    You mention Faulhaber's formula in the title. Have you an idea how you could use it here? – Daniel Fischer Sep 28 '16 at 12:06
  • Unfortunately no, I found this formula by looking for previous similar questions – franchb Sep 28 '16 at 12:08
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    Well. What does Faulhaber's formula say about $1^k + 2^k + \dotsc + n^k$? (You don't need the full formula, by the way, the two most significant terms suffice.) – Daniel Fischer Sep 28 '16 at 12:10
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    Aside, your formula for $x_n$ seems to be wrong. As written, we have $x_n = \frac{n}{k+1} + \frac{3}{2} + o(1)$, that doesn't converge to a finite value. – Daniel Fischer Sep 28 '16 at 12:15
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    That I need to use Bernoulli numbers $Bj$. Also I see $p^(n+1+j)$ in this formula and can't understand what to do with this for my equation – franchb Sep 28 '16 at 12:16
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    Oh well, there isn't an explicit formula for the most significant terms on the wikipedia page. So: $$\sum_{m = 1}^n m^k = \frac{n^{k+1}}{k+1} + \frac{n^k}{2} + \text{ lower order terms}.$$ – Daniel Fischer Sep 28 '16 at 12:23
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    Your expression seems to be divergent. Use integrals (aka upper and lower Riemann sum) for an upper and lower bound of the summation term. Then use squeeze theorem for convergence (or divergence for the expression in your post). – MrYouMath Sep 28 '16 at 12:46
  • If you found a finite limit by simulations and if your simulations are correct then the formula defining $x_n$ in your post is wrong. My guess is that $+\frac{n}{k+n}$ (the current version) should read $-\frac{n}{k+1}$ (but then the limit of $x_n$ shall be $+\frac12$, not $-\frac12$). Please have a careful look. – Did Sep 28 '16 at 13:19
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    Minus $\frac{n}{k+1}$, not plus $\frac{n}{k+1}$. – Did Sep 28 '16 at 13:27
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    Answer accepted and all -- but the question is still wrong. Please correct it. – Did Sep 28 '16 at 14:01
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    No, not done, one still reads the absurd $+\frac{n}{k+1}$. – Did Sep 28 '16 at 14:15

1 Answers1

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Thanks to Faulhaber's formula (and DanielFischer):

$$\sum_{p=1}^np^k=\frac{n^{k+1}}{k+1}+\frac{n^k}2+\mathcal O(n^{k-1})$$

$$x_n=\frac12+\mathcal O(n^{-1})$$

Assuming you meant to say 'minus $\frac{n}{k+1}$'.

Thus, the limit is given as

$$\lim_{n\to\infty}x_n=\frac12+\mathcal O(0)=\frac12$$

if it converges.

Simply Beautiful Art
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    Yes, this is the right limit. Thank you! And thanks to Daniel Fischer - this time I'm solving the equation myself and do the same way. – franchb Sep 28 '16 at 13:44