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

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The discussion centers on comparing the sum of the series $\sum \frac{1}{n^2}$ with the fraction $\frac{61}{36}$. It is established that $\sum \frac{1}{n^2}$ equals $\frac{\pi^2}{6}$, which is approximately 1.64493. The value of $\frac{61}{36}$ is approximately 1.69444. Therefore, $\frac{61}{36}$ is greater than $\sum \frac{1}{n^2}$. The conversation emphasizes the importance of understanding the series' convergence and its value in relation to the fraction.
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$\sum \dfrac{1}{n^2}, \,\, and \,\,\,\dfrac{61}{36}$ which one is bigger ?
 
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Albert said:
$\sum \dfrac{1}{n^2}, \,\, and \,\,\,\dfrac{61}{36}$ which one is bigger ?

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