How to calculate this limit ? $\lim_{n\to\infty}\sum_{k=1}^{2n}(-1)^k\left(\frac{k}{2n}\right)^{100}$ 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/
 A: $$S_{n,m} = \sum_{k=1}^{2n} (-1)^k k^m = -\sum_{k=1}^{2n} k^m + 2 \cdot \sum_{k=1}^{n} (2k)^m$$
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).
A: $$
\lim_{n\to\infty}\sum_{k=0}^{2n}(-1)^k\left(\frac{k}{2n}\right)^{100}
=\lim_{n\to\infty}\sum_{k=0}^{n}\left(\left(\frac{k}{n}\right)^{100}-\left(\frac{k}{n}-\frac1{2n}\right)^{100}\right)
$$
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
$$
\frac12\int_0^1100x^{99}\,\mathrm{d}x=\frac12
$$
