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I ran across the following problem in my Advanced Calculus class:

For a fixed positive number $\beta$, find

$$\lim_{n\to \infty} \left[\frac {1^\beta + 2^\beta + \cdots + n^\beta} {n^{\beta + 1}}\right]$$

I tried manipulating the expression inside the limit but didn't come up with anything useful. I also noted that the numerator can be rewritten as $$\sum_{i=1}^{n}i^\beta$$ which is a well-known formula with a closed form (Faulhaber's formula) but I don't fully understand that formula and we haven't talked about the Bernoulli numbers at all, so I think the author intended for the problem to be solved a different way. Any suggestions on how to tackle this would be much appreciated.

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$$\frac{1^{\beta}+\cdots+n^{\beta}}{n^{\beta+1}}=\frac{1}{n}\Big(\Big(\frac{1}{n}\Big)^{\beta}+\cdots+\Big(\frac{n}{n}\Big)^{\beta}\Big)\to\int_0^1x^{\beta}dx=\frac{1}{\beta+1}.$$

If You did not have integrals yet use Stolz theorem.

  • Excellent, that's exactly what I was looking for. (Sorry for taking so long, I just wanted to make sure I could solve it before I accepted the answer.) Thank you! – Ross Tajvar Jul 14 '14 at 02:34