Albert1
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$\sum \dfrac{1}{n^2}, \,\, and \,\,\,\dfrac{61}{36}$ which one is bigger ?
Albert said:$\sum \dfrac{1}{n^2}, \,\, and \,\,\,\dfrac{61}{36}$ which one is bigger ?
very good solution Kaliprasad !kaliprasad said: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
solution:Albert said: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 ?