find the sum of the series. \n∑(n = 1 to ∞) 8n/((2n - 1)²(2n + 1)²)\nthe sum of the series is .

Explanation:

Step1: Decompose the general - term

First, decompose the general term $\frac{8n}{(2n - 1)^2(2n+1)^2}$ using partial - fraction decomposition. We know that $\frac{8n}{(2n - 1)^2(2n+1)^2}=\frac{1}{(2n - 1)^2}-\frac{1}{(2n+1)^2}$.

Step2: Find the sum of the series

Now, find the sum of the series $\sum_{n = 1}^{\infty}\frac{8n}{(2n - 1)^2(2n+1)^2}=\sum_{n = 1}^{\infty}\left(\frac{1}{(2n - 1)^2}-\frac{1}{(2n+1)^2}\right)$. This is a telescoping series. Let $S_N=\sum_{n = 1}^{N}\left(\frac{1}{(2n - 1)^2}-\frac{1}{(2n+1)^2}\right)$. When $n = 1$, the term is $\frac{1}{1^2}-\frac{1}{3^2}$; when $n = 2$, the term is $\frac{1}{3^2}-\frac{1}{5^2}$; $\cdots$; when $n=N$, the term is $\frac{1}{(2N - 1)^2}-\frac{1}{(2N+1)^2}$. Then $S_N = 1-\frac{1}{(2N + 1)^2}$.

Step3: Find the limit as $N\to\infty$

As $N\to\infty$, $\lim_{N\to\infty}S_N=\lim_{N\to\infty}\left(1-\frac{1}{(2N + 1)^2}\right)$. Since $\lim_{N\to\infty}\frac{1}{(2N + 1)^2}=0$, we have $\lim_{N\to\infty}S_N = 1$.

Answer:

$1$