Sum of 1/n2 or 61/36: Which is Bigger?

[Albert1]

$\sum \dfrac{1}{n^2}, \,\, and \,\,\,\dfrac{61}{36}$ which one is bigger ?

 

[kaliprasad]

$\sum \dfrac{1}{n^2}, \,\, and \,\,\,\dfrac{61}{36}$ which one is bigger ?

Spoiler

assumption n is from 0 to infinite
the 1st term = $\dfrac{\pi^2}{6} \lt \dfrac{10}{6}$
$\dfrac{10}{6} = \dfrac{60}{36} \lt \dfrac{61}{36}$
so second term is bigger

 

[Albert1]

Spoiler

assumption n is from 0 to infinite
the 1st term = $\dfrac{\pi^2}{6} \lt \dfrac{10}{6}$
$\dfrac{10}{6} = \dfrac{60}{36} \lt \dfrac{61}{36}$
so second term is bigger

very good solution Kaliprasad !
suppose $\sum \dfrac {1}{n^2} =\dfrac{\pi^2}{6} $ is not known yet , how can we compare those two numbers ?

 

[Albert1]

very good solution Kaliprasad !
suppose $\sum \dfrac {1}{n^2} =\dfrac{\pi^2}{6} $ is not known yet , how can we compare those two numbers ?

solution:

Spoiler

$\sum \dfrac {1}{n^2}<1+ \dfrac{1}{2^2}+\dfrac {1}{3^2}+\dfrac {1}{3\times 4}+\dfrac {1}{4\times 5}+---+\dfrac {1}{(n-1)\times n}$
$=1+\dfrac {1}{4}+\dfrac{1}{9}+(\dfrac {1}{3}-\dfrac {1}{4})+(\dfrac {1}{4}-\dfrac {1}{5})+------+(\dfrac {1}{n-1}-\dfrac {1}{n})=\dfrac {61}{36}-\dfrac {1}{n}<\dfrac {61}{36}$