Q75 / 170 · Manipal (MET) 2016 · Mathematics

$\lim _{n \rightarrow \infty}\left(\frac{1^2}{1-n^3}+\frac{2^2}{1-n^3}+\ldots+\frac{n^2}{1-n^3}\right)$ is equal to

  • A
    $\frac{1}{3}$
  • Correct answer
    B
    $-\frac{1}{3}$
  • C
    $\frac{1}{6}$
  • D
    $-\frac{1}{6}$

Solution

$\lim _{n \rightarrow \infty} \frac{1}{1-n^3} \sum_{r=1}^n r^2$ $=\lim _{n \rightarrow \infty} \frac{1}{n^3\left(\frac{1}{n^3}-1\right)} \cdot \frac{n(n+1)(2 n+1)}{6}$ $=\lim _{n \rightarrow \infty} \frac{n^3\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{n^3\left(\frac{1}{n^3}-1\right) 6}=-\frac{1}{3}$

Hence, the correct answer is option: 'B'.

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