lim _{n rightarrow infty}left[frac{1^3}{1-n^4 | vedclass

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(D) The given expression is $S_n = \sum_{k=1}^{n} \frac{k^3}{1-n^4}$.Since the denominator $(1-n^4)$ is independent of the summation index $k$,we can

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$-\frac{1}{4}$

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(D) The given expression is $S_n = \sum_{k=1}^{n} \frac{k^3}{1-n^4}$.Since the denominator $(1-n^4)$ is independent of the summation index $k$,we can write:$S_n = \frac{1}{1-n^4} \sum_{k=1}^{n} k^3$.Using the formula for the sum of cubes,$\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}$.Thus,$S_n = \frac{n^2(n+1)^2}{4(1-n^4)}$.Expanding the numerator: $S_n = \frac{n^2(n^2+2n+1)}{4(1-n^4)} = \frac{n^4+2n^3+n^2}{4-4n^4}$.Now,taking the limit as $n \rightarrow \infty$:$\lim _{n \rightarrow \infty} S_n = \lim _{n \rightarrow \infty} \frac{n^4+2n^3+n^2}{4-4n^4}$.Dividing numerator and denominator by $n^4$:$\lim _{n \rightarrow \infty} \frac{1+\frac{2}{n}+\frac{1}{n^2}}{\frac{4}{n^4}-4} = \frac{1+0+0}{0-4} = -\frac{1}{4}$.

$\lim _{n \rightarrow \infty}\left[\frac{1^3}{1-n^4}+\frac{2^3}{1-n^4}+\ldots +\frac{n^3}{1-n^4}\right]=$

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