14th March

Above series is $\sum_{n=-∞}^∞ \dfrac{1}{(nπ^2+1)^2}$ To find answer let's generalise the whole thing. $\sin(x)=x\prod_{n=1}^∞ (1-\frac{x^2}{π^2n^2})$ Taking log on both sides and Differentiating respect to $x$ gives us $\cot(x)= \dfrac{1}{x} +\sum_{n=1}^∞ \dfrac{1}{x+nπ} +\sum_{n=1}^∞ \dfrac{1}{x-nπ}$ $\cot(x)= \sum_{n=-∞}^∞ \dfrac{1}{nπ+x}$ Differentiating respect to $x$ again gives $-\dfrac{1}{\sin^2(x)} = -\sum_{n=-∞}^∞ \dfrac{1}{(nπ+x)^2} \implies \sum_{n=-∞}^∞ \dfrac{1}{(nπ+x)^2}=\dfrac{1}{\sin^2(x)}$ take $x=\dfrac{1}{π}$ It gives $\cdots+\dfrac{1}{(1-2π^2)^2}+\dfrac{1}{(1-π^2)^2} +1 +\dfrac{1}{(1+π^2)^2}+\dfrac{1}{(1+2π^2)^2} +\dfrac{1}{(1+3π^2)^2}+\cdots=\dfrac{1}{π^2 \sin^2(\frac{1}{π})}$ Answer is $\boxed{a=1}$