Question

find $\underset{n\to \infty }{lim}\left(\frac{1}{1-n^2}+\frac{2}{1-n^2}+...+\frac{n}{1-n^2}\right)$

Answer 1

${lim}_{n\to 0}(\frac{1}{1-n^2}+\frac{2}{1-n^2}+....+\frac{n}{1-n^2})={lim}_{n\to 0}\left(\frac{\sum n}{1-n^2}\right)$

$={lim}_{n\to 0}\frac{n(n+1)}{2}\times \frac{1}{(1+n)(1-n)}={lim}_{n\to 0}\left(\frac{n}{2(1-n)}\right)={lim}_{n\to 0}\left(\frac{1}{2(\frac{1}{n}-1)}\right)=-\frac{1}{2}$

$\therefore {lim}_{n\to 0}(\frac{1}{1-n^2}+\frac{2}{1-n^2}+....+\frac{n}{1-n^2})=-\frac{1}{2}$