Question

$\lim_{n \rightarrow \infty}\frac{1}{1^{3} + n^{3}} + \frac{4}{2^{3} + n^{3}} + .... + \frac{1}{2n}$is equal to

• A. $\frac{1}{3}\log_{e}3$
• B. $\frac{1}{3}\log_{e}2$
• C. $\frac{1}{3}\log_{e}\frac{1}{3}$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{3}\log_{e}2$

Answer 2

Let $S = \lim_{n \rightarrow \infty}\frac{1}{1^{3} + n^{3}} + \frac{4}{2^{3} + n^{3}} + ... + \frac{1}{2n}$

$\lim_{x \rightarrow a}\frac{1 - \cos(ax^{2} + bx + c)}{(x - \alpha)^{2}}$

$\therefore S = \lim_{n \rightarrow \infty}\sum_{r = 1}^{n}\frac{r^{2}}{r^{3} + n^{3}} = \lim_{n \rightarrow \infty}\sum_{r = 1}^{n}{}\frac{r^{2}}{n^{3}\left( \frac{r^{3}}{n^{3}} + 1 \right)}$

$= \lim_{n \rightarrow \infty}\sum_{r = 1}^{n}{}\frac{1}{n}.\frac{\left( \frac{r}{n} \right)^{2}}{\left\lbrack 1 + \left( \frac{r}{n} \right)^{3} \right\rbrack}$

Applying the formula, we get

$A = \int_{0}^{1}{}\frac{x^{2}}{1 + x^{3}}dx$

$A = \int_{0}^{1}{}\frac{x^{2}}{1 + x^{3}}dx = \frac{1}{3}\int_{0}^{1}{}\frac{3x^{2}}{1 + x^{3}}dx = \frac{1}{3}\lbrack\log_{e}(1 + x^{3})\rbrack_{0}^{1} = \frac{1}{3}\log_{e}2.$