Q72 / 170 · Manipal (MET) 2017 · Mathematics

$\lim _{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{6 n}\right]$

  • A
    $\log 2$
  • B
    $\log (1+\sqrt{5})$
  • Correct answer
    C
    $\log 6$
  • D
    $0$

Solution

$\begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+5 n}\right] \\ = & \lim _{n \rightarrow \infty} \sum_{r=0}^{5 n} \frac{1}{n+r}=\lim _{n \rightarrow \infty}\left[\frac{1}{n} \sum_{r=0}^{5 n} \frac{n}{n+r}\right] \\ = & \lim _{n \rightarrow \infty}\left[\frac{1}{n} \sum_{r=0}^{5 n} \frac{1}{1+(r / n)}\right] \\ = & \int_0^5 \frac{1}{1+x} d x=[\log (1+x)]_0^5 \\ = & \log 6-\log 1=\log 6\end{aligned}$

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

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