Answer to Question #326682 in Real Analysis for Abdul Wahid

Question

find the nth term of the following series 1*2/3^2*4^2 + 3*4/5^2*6^2 + 5*6/7^2*8^2 + ... is convergent.

Answer 1

ANSWER $\sum_{n=1}^{\infty }\frac{(2n-1)(2n) }{\left ( 2n+1 \right )^{2}\cdot\left ( 2n+2 \right )^{2}}$ converges

EXPLANATION

$a_{n}= \frac{(2n-1)(2n) }{\left ( 2n+1 \right )^{2}\cdot\left ( 2n+2 \right )^{2}} \Rightarrow$

$a_{1} =\frac{(2\cdot1-1)(2\cdot1 ) }{\left ( 2\cdot1+1 \right )^{2}\cdot\left ( 2\cdot1+2 \right )^{2 }} =\frac{1\cdot2}{3^{2}\cdot4^{2}},$ $a_{2} =\frac{(2\cdot2-1)(2\cdot2 ) }{\left ( 2\cdot2+1 \right )^{2}\cdot\left ( 2\cdot2+2 \right )^{2 }} =\frac{3\cdot4}{5^{2}\cdot6^{2}},$ $a_{3} =\frac{(2\cdot3-1)(2\cdot3 ) }{\left ( 2\cdot3+1 \right )^{2}\cdot\left ( 2\cdot3+2 \right )^{2 }} =\frac{5\cdot6}{7^{2}\cdot8^{2}} ...$ .Let $b_{n}= \frac{1}{n^{2}}$ , then $\frac{a_{n}}{b_{n}}=\frac{(2n-1)(2n) \cdot n^{2}}{\left ( 2n+1 \right )^{2}\cdot\left ( 2n+2 \right )^{2}}$ $= \frac{(2 -\frac{1}{n}) \cdot 2 }{\left ( 2 +\frac{1}{n} \right )^{2}\cdot\left ( 2 +\frac{2}{n} \right )^{2}}$ . Since $\lim _{n\rightarrow\infty}\frac{1}{n}=0$ , then $\lim _{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=\frac{1}{4}>0.$

The series $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ converges (p-series, p=2). Hence (by Limit Comparison Test) $\sum_{n=1}^{\infty }\frac{(2n-1)(2n) }{\left ( 2n+1 \right )^{2}\cdot\left ( 2n+2 \right )^{2}}$ converges.