Question

$n\stackrel{Lt}{\to }\infty [\frac{1}{1-n^2}+\frac{2}{1-n^2}+\dots +\frac{n}{1-n^2}]=$

• A. 0
• B. -1/2
• C. 1/2
• D. 1

Answer 1

Answer: (B)

-1/2

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