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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?

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    $\begingroup$ A is a diverging series its $\sum \frac{1}{n} $ while $\sum \frac{1}{n^2}$ is a convergent series $\endgroup$ Commented Feb 18, 2020 at 17:11
  • $\begingroup$ What is the $T_n$ term of a series? $\endgroup$
    – jijijojo
    Commented Feb 18, 2020 at 17:26

1 Answer 1

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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 $$\frac{A}4$$

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