show that

9/8<summation(1/(n^3))<5/4

2. Originally Posted by nzm88
show that

9/8<summation(1/(n^3))<5/4

Do you mean $\sum_{n=1}^{\infty} \frac{1}{n^3}$ is bounded between $\frac{9}{8}$ and $\frac{5}{4}$ ?

Consider $\sum_{n=1}^{\infty} \frac{1}{n^3}= 1 + \frac{1}{2^3} + \frac{1}{3^3} + ... > 1 + \frac{1}{2^3 } = 1 + \frac{1}{8} = \frac{9}{8}$ , the lower bound .

Now , we are going to find the upper bound :

$\sum_{n=1}^{\infty} \frac{1}{n^3}$

$= 1 + \sum_{n=2}^{\infty} \frac{1}{n^3}$

Consider $0 < n^3 - n < n^3 , n \geq 2$

$\frac{1}{ n^3 - n} > \frac{1}{n^3}$

$\sum_{n=2}^{\infty}\frac{1}{ n^3 - n} > \sum_{n=2}^{\infty}\frac{1}{n^3}$

$\sum_{n=2}^{\infty} \left [ \frac{1}{n(n^2 - 1)} \right ] > \sum_{n=2}^{\infty}\frac{1}{n^3}$

$\sum_{n=2}^{\infty} \left [ \frac{1}{2} \left ( \frac{1}{ n -1} + \frac{1}{n + 1} \right ) - \frac{1}{n} \right ] > \sum_{n=2}^{\infty}\frac{1}{n^3}$

The left hand side is a telescoping series , the value is $\frac{1}{4}$

Therefore ,

$\frac{1}{4} > \sum_{n=2}^{\infty}\frac{1}{n^3}$

$1 + \frac{1}{4 } = \frac{5}{4} > 1 + \sum_{n=2}^{\infty}\frac{1}{n^3} = \sum_{n=1}^{\infty}\frac{1}{n^3}$

3. Originally Posted by simplependulum
Do you mean $\sum_{n=1}^{\infty} \frac{1}{n^3}$ is bounded between $\frac{9}{8}$ and $\frac{5}{4}$ ?

Consider $\sum_{n=1}^{\infty} \frac{1}{n^3}= 1 + \frac{1}{2^3} + \frac{1}{3^3} + ... > 1 + \frac{1}{2^3 } = 1 + \frac{1}{8} = \frac{1}{9}$ , the lower bound .
Typo: $1+ \frac{1}{8}= \frac{8}{9}$, of course.

Now , we are going to find the upper bound :

$\sum_{n=1}^{\infty} \frac{1}{n^3}$

$= 1 + \sum_{n=2}^{\infty} \frac{1}{n^3}$

Consider $0 < n^3 - n < n^3 , n \geq 2$

$\frac{1}{ n^3 - n} > \frac{1}{n^3}$

$\sum_{n=2}^{\infty}\frac{1}{ n^3 - n} > \sum_{n=2}^{\infty}\frac{1}{n^3}$

$\sum_{n=2}^{\infty} \left [ \frac{1}{n(n^2 - 1)} \right ] > \sum_{n=2}^{\infty}\frac{1}{n^3}$

$\sum_{n=2}^{\infty} \left [ \frac{1}{2} \left ( \frac{1}{ n -1} + \frac{1}{n + 1} \right ) - \frac{1}{n} \right ] > \sum_{n=2}^{\infty}\frac{1}{n^3}$

The left hand side is a telescoping series , the value is $\frac{1}{4}$

Therefore ,

$\frac{1}{4} > \sum_{n=2}^{\infty}\frac{1}{n^3}$

$1 + \frac{1}{4 } = \frac{5}{4} > 1 + \sum_{n=2}^{\infty}\frac{1}{n^3} = \sum_{n=1}^{\infty}\frac{1}{n^3}$

5. Originally Posted by HallsofIvy
Typo: $1+ \frac{1}{8}= \frac{8}{9}$, of course.
Is this another typo?
its $\frac{9}{8}$.

6. Originally Posted by TWiX
Is this another typo?
its $\frac{9}{8}$.
fractional dyslexia ... many are afflicted.

7. Originally Posted by skeeter
fractional dyslexia ... many are afflicted.
Do not use hard words
Use simple words
My English is horrible

8. HaHa , I am just wondering what if my post is published .