Definite Integration Question 21
Question 21 - 01 February - Shift 1
$\lim _{n \to \infty}(\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots+\frac{1}{2 n})$ is equal to :-
(1) 0
(2) $\log _e 2$
(3) $\log _e(\frac{3}{2})$
(4) $\log _e(\frac{2}{3})$
Show Answer
Answer: (2)
Solution:
Formula: Definite integrals as a limit of sum, Standard formulas for Indefinite Integration
$ \begin{aligned} &L= \lim _{n \to \infty}(\frac{1}{1+n}+\ldots+\frac{1}{n+n})\\ & L=\lim _{n \to \infty} \sum _{r=1}^{n} \frac{1}{n+r} \\ & L=\lim _{n \to \infty} \sum _{r=1}^{n} \frac{1}{n}(\frac{1}{1+\frac{r}{n}}) \\ & L=\int_0^{1} \frac{1}{1+x} d x=[\ln (1+x]_0^{1}=\ln 2 \\ & L = \ln2 =log _e2 \end{aligned} $
