Let-S-n-1-1-3-1-2-1-3-2-3-1-2-n-1-3-2-3-n-3-n-1-2-3-Then-S-n-is-greater-than-

Question Number 27604 by 0722136841 last updated on 10/Jan/18

Let

S_{n} = \frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + \dots + \frac{1 + 2 + \dots + n}{1^{3} + 2^{3} + \dots + n^{3}}; n = 1, 2, 3, . .

Then S_{n} is greater than

Commented by abdo imad last updated on 11/Jan/18

S_{n} = \sum_{k = 1}^{n} \frac{\sum_{p = 1}^{k} p}{\sum_{p = 1}^{k} p^{3}} l e t s i m p l i f y S_{n}

w e k n o w t h a t \sum_{p = 1}^{k} p = \frac{k (k + 1)}{2} a n d \sum_{p = 1}^{k} p^{3} = {(\frac{k (k + 1)}{2})}^{2}

S_{n} = \sum_{k = 1}^{n} \frac{2}{k (k + 1)} = 2 \sum_{k = 1}^{n} (\frac{1}{k} - \frac{1}{k + 1})

S_{n} = 2 (1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \dots + \frac{1}{n} - \frac{1}{n + 1}) = 2 (1 - \frac{1}{n + 1})

= \frac{2 n}{n + 1} b u t n + 1 < 2 n f o r a l l n > 1 \Rightarrow \frac{1}{n + 1} > \frac{1}{2 n}

\Rightarrow \frac{2 n}{n + 1} > \frac{1}{2} \Rightarrow S_{n} > \frac{1}{2} a n d a l s o w e h a v e \frac{1}{2} < S_{n} < 2

a n d {l i m}_{n - >\propto} S_{n} = 2