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An elegant result on sum of powers

2010年02月21日 留下评论 Go to comments

\displaystyle\lim_{m\rightarrow \infty}\left[\left(\frac{1}{m}\right)^m+\left(\frac{2}{m}\right)^m+\cdots+\left(\frac{m-1}{m}\right)^m\right]=\frac{1}{e-1}

Here it is. It is highly non-trivial that the left hand side, involving sum of powers of integers, yields something like e. Elegant isn’t it.  To prove it you need Euler-Maclaurin formula which is said to be one of the indispensable tools in any mathematitian’s bag of tricks but not taught in any undergraduate course – regardless of its elementary nature, nothing more advanced than calculus. The proof can be found here

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