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?

As an alternative
$$n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + \ldots \right )=$$
$$=n\left(\frac12 \frac1n+\frac12 \frac1n- \frac{1}{n+1} + \frac12\frac{1}{n+2}+\frac12\frac{1}{n+2}-\frac{1}{n+3}+\frac12\frac{1}{n+4}+\ldots\right)=$$
$$=\frac12+n\sum_{k=0}^\infty \frac{1}{(n+2k)(n+2k+1)(n+2k+2)} \to \frac12$$
indeed
$$n\sum_{k=0}^\infty \frac{1}{(n+2k)(n+2k+1)(n+2k+2)} \le n\sum_{k=0}^\infty \frac{1}{(n+2k)^3} =$$
$$=\frac1n\int_0^\infty \frac1{(1+2x)^3}dx=\frac 1{4n} \to 0$$
