Let $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}.... =A$. What is $\frac {1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}....$ in terms of A?

Question

Find the value of $\frac {1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}....\infty$ in terms of A

For the first expression, the $T_n$ term will be $$T_n=\frac{1}{\frac{((n)(n+1))^2}{4}}$$

Now $$T_n=4\left[\sum \frac{1}{(n)^2(n+1)^2}\right]$$

How should I proceed?

Answer 1

Given expression$$\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}…$$

Which is equal to $$\frac1{2^2}\left[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}…\right]$$

Hence the given Series sums up to $$\frac{A}4$$