Let $S_{n} = \frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + ... + \frac{1 + 2 + ... + n}{1^{3} + 2^{3} + ... + n^{3}},\ n = 1,2,3,\ldots$. Then $S_{n}$ is not greater than

Question

Dr. Abubacker · Accepted Answer

(c)
Here, $T_{n} = \frac{1 + 2 + 3 + ... + n}{1^{3} + 2^{3} + 3^{3} + ... + n^{3}} = \frac{\Sigma n}{\Sigma n^{3}}$
$= \frac{n(n + 1)/2}{\left\{ n(n + 1)/2 ight\}^{2}} = \frac{2}{n(n + 1)}$
$= 2\left( \frac{1}{n} - \frac{1}{n + 1} ight)$
$	herefore S_{n} = \Sigma T_{n}$
$= 2\left( \frac{1}{1} - \frac{1}{2} ight) + 2\left( \frac{1}{2} - \frac{1}{3} ight) + ... + 2\left( \frac{1}{n} - \frac{1}{n + 1} ight)$
$= 2\left( 1 - \frac{1}{n + 1} ight) = 2 - \frac{2}{n + 1} \leq 2\ \ \ \ \ \left\lbrack \because\frac{2}{n + 1} \leq 1 ightbrack$

Q. Let $S_{n} = \frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + ... + \frac{1 + 2 + ... + n}{1^{3} + 2^{3} + ... + n^{3}},\ n = 1,2,3,\ldots$. Then $S_{n}$ is not greater than

Explanation

Reviewed by Dr. Abubacker

Want more practice?