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Question
\[\lim_{n \to \infty} \left\{ \frac{1}{1 - n^2} + \frac{2}{1 - n^2} + . . . + \frac{n}{1 - n^2} \right\}\]
Options
(a) 0
(b) −1/2
(c) 1/2
(d) none of these
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Solution
(b) \[\frac{- 1}{2}\]
\[\lim_{n \to \infty} \left[ \frac{1}{1 - n^2} + \frac{2}{1 - n^2} + . . . . . . + \frac{n}{1 - n^2} \right]\]
\[ = \lim_{n \to \infty} \left( \frac{1}{1 - n^2} \right) \left[ 1 + 2 + 3 . . . . . n \right]\]
\[ = \lim_{n \to \infty} \left( \frac{1}{1 - n^2} \right) \left[ \frac{n\left( n + 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{1}{2}\left[ \frac{n\left( n + 1 \right)}{\left( 1 - n \right) \left( 1 + n \right)} \right]\]
\[ = \lim_{n \to \infty} \frac{1}{2} \frac{n}{1 - n}\]
\[ = \lim_{n \to \infty} \frac{1}{2} \frac{1}{\frac{1}{n} - 1}\]
\[ = \frac{1}{2}\left( - 1 \right)\]
\[ = \frac{- 1}{2}\]
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