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$$ \lim_{n \to \infty} \frac{1^m + 2^m + 3^m + \cdots + (2n-1)^m }{n^{m+1}} $$

I am kind of stuck since I cannot make it look into a form that would involve the integral of certain function. I know somehow it would be easy if we can compare this limit to a Riemann sum. Any ideas?

Another User
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  • See http://math.stackexchange.com/questions/465075/find-lim-limits-n-to-infty-frac1n-sum-limits2n-r-1-fracr-sq and http://math.stackexchange.com/questions/478344/what-is-the-result-of-lim-n-to-infty-frac-sumn-i-1-iknk1?noredirect=1&lq=1 – lab bhattacharjee Nov 02 '16 at 06:42
  • Notice sum and integral are asymptotic of each other. Both are polynomials so this implies their main terms are equal. This justifies the answers below ; the limit must exist and must be the one by the integral. This is sketchy ofcourse but the main idea. – mick May 19 '23 at 18:15

2 Answers2

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Observe \begin{align} \sum^{2n-1}_{k=1} \frac{k^m}{n^m n} = \sum^{2n-1}_{k=1} \left(\frac{k}{n} \right)^m\frac{1}{n} \approx \int^2_0 x^m\ dx. \end{align}

Jacky Chong
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16 years later: More details (see Jacky's answer) $$ \frac{1^m + 2^m + 3^m + \cdots + (2n-1)^m }{n^{m+1}} =\frac{1}{n} \left[\left(\frac{1}{n}\right)^m +\left(\frac{2}{n}\right)^m +\left(\frac{3}{n}\right)^m + \cdots +\left(\frac{2n-1}{n}\right)^m \right] $$ Recognize this as a Riemann sum for $\int_0^2 x^m\;dx$, where the interval $[0,2]$ is subdivided into $2n$ equal-size intervals, and the function is evaluated at the left endpoint of each interval. Then cite appropriate theorems to show (a) that $x^m$ is integrable on $[0,1]$, and (b) these Riemann sums converge to the integral.

Remark. In this proof, the value $m$ could be any nonnegative real number. Or even any complex number with nonnegative real part.

The integral converges for $-1<m<0$, but the elementary reasoning here will not show that the Riemann sums converge to the integral.

GEdgar
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  • +1 Nice to answer this old question. Although I think it can be improved, it is already a big improvement. I think it can be shown in a different way, with a difference operator. – mick May 19 '23 at 18:07