<s>[INST] Can we find $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) $? I have got one method,
If we consider $ a_{n} = \int_{0}^{1} \frac{nx^{n-1}}{1+x} \ dx $
Then, $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) = \lim_{n \to \infty }a_{n} = \frac{1}{2} $
But can anyone attack this problem in a different & more standard way?
 [/INST] We have that
$$H_N=\sum_{k=1}^{N} \frac{1}k=\ln N+\gamma+\frac1{2N}+O\left(\frac1{N^2}\right)$$
then
$$\sum_{k=1}^{2N} \frac{(-1)^{k+1}}k=H_{2N}-H_{N}=\log {2}-\frac1{2N}+O\left(\frac1{N^2}\right)$$
and
$$\sum_{k=n}^{2N} \frac{(-1)^{k+1}}k=\sum_{k=1}^{2N} \frac{(-1)^{k+1}}k-\sum_{k=1}^{n-1} \frac{(-1)^{k+1}}k=$$
$$=-\frac1{2N}+O\left(\frac1{N^2}\right)+\frac1{2(n-1)}+O\left(\frac1{n^2}\right) \sim \frac1{2(n-1)}+O\left(\frac1{n^2}\right)$$
then
$$n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + \ldots \right )\sim \frac n{2(n-1)}+O\left(\frac1{n}\right) \to \frac12$$
</s>