Find the sum of $n^{th}$ term of the series: $$\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+\cdots$$
I could not find the rule for the $n^{th}$ term.
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$\begingroup$ Is it infinite or just $3$ terms as dots are missing? $\endgroup$– Archis WelankarMar 13, 2016 at 11:31
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$\begingroup$ @Archis, its infinite. $\endgroup$– Iaamuser userMar 13, 2016 at 11:36
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$\begingroup$ @Iaamuseruser If it is infinite, what do you mean by the $n^{th}$ term? $\endgroup$– GoodDeedsMar 13, 2016 at 11:41
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$\begingroup$ @GoodDeeds, I have typed the question as it is given. $\endgroup$– Iaamuser userMar 13, 2016 at 11:47
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1 Answer
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Let the $n^{th}$ of the corresponding sequence be $a_n$. $$a_n=\frac{2n+1}{n^2(n+1)^2}=\frac1{n^2}-\frac1{(n+1)^2}$$ Thus, the sum $S_n$, telescopes and is
$$S_n=1-\frac14+\frac14-\frac19+-\cdots+\frac1{n^2}-\frac1{(n+1)^2}=1-\frac1{(n+1)^2}$$
answered Mar 13, 2016 at 11:31