Answer to Question #264590 in Real Analysis for Nikhil

Question

Test the following series for convergence

2.3/(4^2.5^2)+ 4.5/6^2.7^2) + 6.7/ (8^2.9^2) ..+.......

Answer 1

Let us test the series $\sum\limits_{n=1}^{\infty}\frac{2n(2n+1)}{(2n+2)^2(2n+3)^2}$ for convergence. Let us consider the series $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}.$

Since $\lim\limits_{n\to\infty}\frac{\frac{2n(2n+1)}{(2n+2)^2(2n+3)^2}}{\frac{1}{n^2}} =\lim\limits_{n\to\infty}\frac{2n^3(2n+1)}{(2n+2)^2(2n+3)^2} =\lim\limits_{n\to\infty}\frac{4+\frac{2}{n}}{(2+\frac{2}n)^2(2+\frac{3}n)^2} =\frac{4}{2^22^2}=\frac{1}4,$

we conclude that the series $\sum\limits_{n=1}^{\infty}\frac{2n(2n+1)}{(2n+2)^2(2n+3)^2}$ is equivalent to the series $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}$. Since the $s$-series $\sum\limits_{n=1}^{\infty}\frac{1}{n^s}$ is covergent for $s>1,$ we conclude that the series $\sum\limits_{n=1}^{\infty}\frac{2n(2n+1)}{(2n+2)^2(2n+3)^2}$ is covergent as well.