MCQ
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots+\frac{n}{n^2}\right)$ is
  • $\frac{1}{2}$
  • B
  • C
    1
  • D
    $-\frac{1}{2}$

Answer

Correct option: A.
$\frac{1}{2}$
(A)
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots+\frac{n}{n^2}\right)$
$=\lim _{n \rightarrow \infty}\left(\frac{1+2+3+\ldots+n}{n^2}\right)=\lim _{n \rightarrow \infty} \frac{\frac{n(n+1)}{2}}{n^2}$
$=\frac{1}{2} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=\frac{1}{2}$
Alternate method:
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots .+\frac{n}{n^2}\right)$
$=\lim _{n \rightarrow \infty}\left(\frac{1+2+3+\ldots+n}{n^2}\right)$
$\lim _{x \rightarrow \infty} \frac{1^\alpha+2^\alpha+3^\alpha+\ldots+x^\alpha}{x^{\alpha+1}}=\frac{1}{\alpha+1}$
$=\frac{1}{1+1} $
$=\frac{1}{2}$

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