_ n [ 1 n + 1 n + 1 + 1 n + 2 + .....1 2n ] = - Vidyadip

MCQ

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + .....\frac{1}{{2n}}} \right] = $

A
$0$
B
${\log _e}4$
C
${\log _e}3$
✓
${\log _e}2$

✓

Answer

Correct option: D.

${\log _e}2$

d
(d) $\mathop {{\rm{lim}}}\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + ..... + \frac{1}{{2n}}} \right]$

$= \mathop {{\rm{lim}}}\limits_{n \to \infty } \,\left[ {\frac{1}{n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + .... + \frac{1}{{n + n}}} \right]$

$ = \frac{1}{n}\mathop {{\rm{lim}}}\limits_{n \to \infty } \left[ {1 + \frac{1}{{1 + \frac{1}{n}}} + \frac{1}{{1 + \frac{2}{n}}} + .... + \frac{1}{{1 + \frac{n}{n}}}} \right]$

$ = \frac{1}{n}\mathop {{\rm{lim}}}\limits_{n \to \infty } \sum\limits_{r = 0}^n {\left[ {\frac{1}{{1 + \frac{r}{n}}}} \right]} $ $ = \int_0^1 {\frac{1}{{1 + x}}\,\,dx} $

$ = [{\log _e}(1 + x)]_0^1 = {\log _e}2 - {\log _e}1 $

$= {\log _e}2$.

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