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?

This would be my "napkin" heuristic:
Since $\left(\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \cdots\right)$ is the absolute value of the tail of a convergent series, it tends to zero.  Therefore,
$$\begin{align*}\limsup_{n\to\infty} n&\left(\frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \cdots\right) \\ = 1 &- \liminf_{n\to\infty}\, (n+1)\left(\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \cdots\right) \\ &+\lim_{n\to\infty}\left(\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \cdots\right) \\ = 1 &- \liminf_{n\to\infty} n\left(\frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \cdots\right)\end{align*}$$
From here, we find that if the limit in question exists, it must equal $\frac{1}{2}.$
