It incorrectly evaluates $\displaystyle\sum_{n=0}^{\infty}\frac{1}{((2n+1)^2-4)^2}=\frac{\pi^2}{64}-\frac{1}{12}$, but correct answer is $\displaystyle\frac{\pi^2}{64}$
![]() | 1 | initial version |
It incorrectly evaluates $\displaystyle\sum_{n=0}^{\infty}\frac{1}{((2n+1)^2-4)^2}=\frac{\pi^2}{64}-\frac{1}{12}$, but correct answer is $\displaystyle\frac{\pi^2}{64}$
![]() | 2 | retagged |
It incorrectly evaluates $\displaystyle\sum_{n=0}^{\infty}\frac{1}{((2n+1)^2-4)^2}=\frac{\pi^2}{64}-\frac{1}{12}$, but correct answer is $\displaystyle\frac{\pi^2}{64}$
![]() | 3 | retagged |
It incorrectly evaluates $\displaystyle\sum_{n=0}^{\infty}\frac{1}{((2n+1)^2-4)^2}=\frac{\pi^2}{64}-\frac{1}{12}$, but correct answer is $\displaystyle\frac{\pi^2}{64}$
![]() | 4 | retagged |
It incorrectly evaluates $\displaystyle\sum_{n=0}^{\infty}\frac{1}{((2n+1)^2-4)^2}=\frac{\pi^2}{64}-\frac{1}{12}$, but correct answer is $\displaystyle\frac{\pi^2}{64}$