MCQ
$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{1^3} + {n^3}}} + \frac{4}{{{2^3} + {n^3}}} + .... + \frac{1}{{2n}}$ is equal to
- A$\frac{1}{3}{\log _e}3$
- ✓$\frac{1}{3}{\log _e}2$
- C$\frac{1}{3}{\log _e}\frac{1}{3}$
- DNone of these
✓
Answer
Correct option: B.
$\frac{1}{3}{\log _e}2$
b
(b) Let $S = \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{{{1^3} + {n^3}}} + \frac{4}{{{2^3} + {n^3}}} + ... + \frac{1}{{2n}}$
(b) Let $S = \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{{{1^3} + {n^3}}} + \frac{4}{{{2^3} + {n^3}}} + ... + \frac{1}{{2n}}$
$S = \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{{{1^3} + {n^3}}} + \frac{4}{{{2^3} + {n^3}}} + ... + \frac{n^2}{{n^3 + n^3}}$
$\therefore \,\,\,S = \mathop {\lim }\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^n {\,\,\frac{{{r^2}}}{{{r^3} + {n^3}}}} $
$ = \mathop {\lim }\limits_{n \to \infty } \,\,\,\sum\limits_{r = 1}^n {} \frac{{{r^2}}}{{{n^3}\,\left( {\frac{{{r^3}}}{{{n^3}}} + 1} \right)}}$
$ = \mathop {\lim }\limits_{n \to \infty } \,\,\,\sum\limits_{r = 1}^n {} \frac{1}{n}.\frac{{{{\left( {\frac{r}{n}} \right)}^2}}}{{\left[ {1 + \,{{\left( {\frac{r}{n}} \right)}^3}} \right]}}$
Applying the formula, we get
$A = \int_0^1 {} \frac{{{x^2}}}{{1 + {x^3}}}dx$
$ \frac{1}{3}\int_0^1 {} \frac{{3{x^2}}}{{1 + {x^3}}}dx$
$ = \frac{1}{3}[{\log _e}(1 + {x^3})]_0^1 = \frac{1}{3}{\log _e}2.$
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