Answer to Question #272701 in Real Analysis for Sourav

Question #272701

Is (3x4)/5²+(5x6)/7²+(7x8)/9² +..... series convergence or not

1
Expert's answer
2021-11-29T15:48:28-0500
3(4)52+5(6)72+7(8)92+...=n=1(2n+1)(2n+2)(2n+3)2\dfrac{3(4)}{5^2}+\dfrac{5(6)}{7^2}+\dfrac{7(8)}{9^2}+...=\displaystyle\sum_{n=1}^{\infin}\dfrac{(2n+1)(2n+2)}{(2n+3)^2}

an=(2n+1)(2n+2)(2n+3)2a_n=\dfrac{(2n+1)(2n+2)}{(2n+3)^2}

limnan=limn(2n+1)(2n+2)(2n+3)2\lim\limits_{n\to\infin}a_n=\lim\limits_{n\to\infin}\dfrac{(2n+1)(2n+2)}{(2n+3)^2}

=limn(2n/n+1/n)(2n/n+2/n)(2n/n+3/n)2=\lim\limits_{n\to\infin}\dfrac{(2n/n+1/n)(2n/n+2/n)}{(2n/n+3/n)^2}

=limn(2+1/n)(2+2/n)(2+3/n)2=\lim\limits_{n\to\infin}\dfrac{(2+1/n)(2+2/n)}{(2+3/n)^2}

=(2+0)(2+0)(2+0)2=10=\dfrac{(2+0)(2+0)}{(2+0)^2}=1\not=0

Then the series


3(4)52+5(6)72+7(8)92+...=n=1(2n+1)(2n+2)(2n+3)2\dfrac{3(4)}{5^2}+\dfrac{5(6)}{7^2}+\dfrac{7(8)}{9^2}+...=\displaystyle\sum_{n=1}^{\infin}\dfrac{(2n+1)(2n+2)}{(2n+3)^2}

diverges by the Test for Divergence.



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