Question
$$\mathop {\lim }\limits_{n\, \to \,\infty } \left\{ {\frac{1}{{1 - {n^2}}} + \frac{2}{{1 - {n^2}}} + ..... + \frac{n}{{1 - {n^2}}}} \right\}$$ is equal to-
A.
$$0$$
B.
$$ - \frac{1}{2}$$
C.
$$ \frac{1}{2}$$
D.
none of these
Answer :
$$ - \frac{1}{2}$$
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{n\, \to \,\infty } \left\{ {\frac{1}{{1 - {n^2}}} + \frac{2}{{1 - {n^2}}} + ..... + \frac{n}{{1 - {n^2}}}} \right\} \cr
& = \mathop {\lim }\limits_{n\, \to \,\infty } \frac{{1 + 2 + 3 + ..... + n}}{{1 - {n^2}}} \cr
& = \mathop {\lim }\limits_{n\, \to \,\infty } \,\frac{{\sum n }}{{1 - {n^2}}} \cr
& = \mathop {\lim }\limits_{n\, \to \,\infty } \,\frac{{n\left( {n + 1} \right)}}{{2\left( {1 - {n^2}} \right)}} \cr
& = \mathop {\lim }\limits_{n\, \to \,\infty } \frac{{1 + \frac{1}{n}}}{{2\left[ {\frac{1}{{{n^2}}} - 1} \right]}} = - \frac{1}{2} \cr} $$