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How to calculate this limit?

$$\lim_{n\to\infty}\sum_{k=1}^{2n}(-1)^k\left(\frac{k}{2n}\right)^{100}$$ use integration but difficult/

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What do yo umean by “using integrate but difficult”? Is this from an exercise which requires that you use integration? –  Harald Hanche-Olsen Feb 11 '13 at 7:59
    
The term alternation makes Riemann sums a problem. –  robjohn Feb 11 '13 at 18:56

2 Answers 2

$$S_{n,m} = \sum_{k=1}^{2n} (-1)^k k^m = -\sum_{k=1}^{2n} k^m + 2 \cdot \sum_{k=1}^{n} (2k)^m$$ Now recall that $$\sum_{k=1}^n k^m = \dfrac{n^{m+1}}{m+1} + \dfrac{n^m}2 + \mathcal{O}(n^{m-1})$$ Hence, \begin{align} S_{n,m} & = - \dfrac{(2n)^{m+1}}{m+1} - \dfrac{(2n)^m}2 + \mathcal{O}(n^{m-1}) + 2^{m+1} \left(\dfrac{n^{m+1}}{m+1} + \dfrac{n^m}2 + \mathcal{O}(n^{m-1})\right)\\ & = - 2^{m-1} n^m + 2^m n^m + \mathcal{O}(n^{m-1})\\ & = 2^{m-1} n^m + \mathcal{O}(n^{m-1}) \end{align} The limit you are interested in is $$\lim_{n \to \infty} \dfrac{S_{n,m}}{(2n)^{m}} = \lim_{n \to \infty} \dfrac{2^{m-1} n^m + \mathcal{O}(n^{m-1})}{2^m n^m} = \dfrac12 + \lim_{n \to \infty}\mathcal{O} \left(\dfrac1n\right) = \dfrac12$$ Note that the limit is $1/2$ independent of $m$ (which is $100$ in your case).

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I was going to add a second part to my answer using the Euler-Maclaurin Sum Formula, but now I won't bother (+1). –  robjohn Feb 11 '13 at 19:24

$$ \lim_{n\to\infty}\sum_{k=0}^{2n}(-1)^k\left(\frac{k}{2n}\right)^{100} =\lim_{n\to\infty}\sum_{k=0}^{n}\left(\left(\frac{k}{n}\right)^{100}-\left(\frac{k}{n}-\frac1{2n}\right)^{100}\right) $$ By the Mean Value Theorem, this is between $$ \lim_{n\to\infty}\sum_{k=0}^{n}100\left(\frac{k}{n}\right)^{99}\frac1{2n}\quad\text{and}\quad\lim_{n\to\infty}\sum_{k=0}^{n}100\left(\frac{k}{n}-\frac1{2n}\right)^{99}\frac1{2n} $$ which are both Riemann sums for $$ \frac12\int_0^1100x^{99}\,\mathrm{d}x=\frac12 $$

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+1. Ha ha. I wanted to write the Riemann sum approach and saw you had already done. Hence, had to write a different approach :). –  user17762 Feb 11 '13 at 21:44

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