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?

We have:
\begin{align}
n\left (\frac{1}{n} - \frac{1}{n+1} + \cdots \right ) &= n \left ( \frac{1}{n(n+1)} + \frac{1}{(n+2)(n+3)} + \cdots \right ) \\
&\le n \left ( \frac{1}{n^2} + \frac{1}{(n+2)^2} + \cdots \right ) \\
&\le n \int_{n-2}^{\infty} \frac{1}{2x^2}dx = \frac{n}{2(n-2)}
\end{align}
Similarly,
\begin{align}
n\left (\frac{1}{n} - \frac{1}{n+1} + \cdots \right ) \ge \frac{n}{2(n+1)}
\end{align}
Thus, by letting $n$ tends to infinity, we obtain
\begin{align}
\lim_{n\to \infty} {n \left ( \frac{1}{n} - \frac{1}{n+1} + \cdots \right )} = \frac{1}{2}
\end{align}
