1. ## Series

$b_{n} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\f rac{!}{n^2}$

a) Prove that this series have limit b ,
b) prove that the limit b , hold $\frac{49}{36} < b < 2$

I start to solve :

$\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac {1}{n-1} - \frac{1}{n}$

so
$b_{n} = 1 + \frac{1}{1}-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... \frac{1}{n-1} - \frac{1}{n} = 2 - \frac{1}{n} < 2$

How I prove that the series rise ?

and how I prove b)

2. Originally Posted by gilyos
$b_{n} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\f rac{!}{n^2}$

a) Prove that this series have limit b ,
b) prove that the limit b , hold $\frac{49}{36} < b < 2$

I start to solve :

$\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac {1}{n-1} - \frac{1}{n}$

so
$b_{n} = 1 + \frac{1}{1}-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... \frac{1}{n-1} - \frac{1}{n} = 2 - \frac{1}{n} < 2$

How I prove that the series rise ?

and how I prove b)
O' right , you have found the upper bound ,

the lower bound , umm ... What you have to do actually depends on what the answer is ..

Consider the series

$S = a_1 + a_2 + a_3 + a_4 + ...$

we know any other series with a cancel of any members $a_k$ must be less than $S$, provided $a_k > 0$ .

Therefore ,

$1 + \frac{1}{2^2 } + \frac{1}{3^2} + \frac{ 1}{4^2} + ... > 1 + \frac{1}{2^2} + \frac{1}{3^2} = 1 + \frac{1}{4} + \frac{1}{9} = \frac{ 36 + 9 + 4 }{36} = \frac{49}{36}$