please help me solve this equation

**nzm88** · January 31st 2010, 01:34 AM

show that

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

**simplependulum** · January 31st 2010, 02:28 AM

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 ,

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

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

**HallsofIvy** · January 31st 2010, 02:43 AM

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 ,

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

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

**nzm88** · January 31st 2010, 05:06 AM

thanx for your answer. i really appreciate it

**TWiX** · January 31st 2010, 05:09 AM

Originally Posted by HallsofIvy

Typo: $1+ \frac{1}{8}= \frac{8}{9}$ , of course.

Is this another typo?

its $\frac{9}{8}$ .

**skeeter** · January 31st 2010, 05:22 AM

Originally Posted by TWiX

Is this another typo?

its $\frac{9}{8}$ .

fractional dyslexia ... many are afflicted.

**TWiX** · January 31st 2010, 05:34 AM

Originally Posted by skeeter

fractional dyslexia ... many are afflicted.

Do not use hard words

Use simple words

My English is horrible

**simplependulum** · January 31st 2010, 10:58 PM

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

.

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