The value of lim_{n to infty} [frac{1}{1 - n^ | vedclass

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(D) The given expression is $S_n = \sum_{k=1}^{n} \frac{k}{1 - n^2}$.Since the denominator $(1 - n^2)$ is independent of $k$, we can write $S_n = \fra

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$-\frac{1}{2}$

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(D) The given expression is $S_n = \sum_{k=1}^{n} \frac{k}{1 - n^2}$.Since the denominator $(1 - n^2)$ is independent of $k$, we can write $S_n = \frac{1}{1 - n^2} \sum_{k=1}^{n} k$.Using the formula for the sum of the first $n$ natural numbers, $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.Thus, $S_n = \frac{n(n+1)}{2(1 - n^2)}$.We can factor the denominator: $1 - n^2 = -(n^2 - 1) = -(n-1)(n+1)$.So, $S_n = \frac{n(n+1)}{-2(n-1)(n+1)} = \frac{n}{-2(n-1)}$.Now, take the limit as $n 	o \infty$:$\lim_{n 	o \infty} \frac{n}{-2n + 2} = \lim_{n 	o \infty} \frac{n}{n(-2 + \frac{2}{n})} = \frac{1}{-2 + 0} = -\frac{1}{2}$.

The value of $\lim_{n \to \infty} [\frac{1}{1 - n^2} + \frac{2}{1 - n^2} + \dots + \frac{n}{1 - n^2}]$ is

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