take home for 4/8 converges or diverges? m8 w=1 (2w + 1)(w - 1) (w+1)^2

Explanation:

Step1: Apply the ratio - test

Let (a_n=\frac{(2n + 1)(n - 1)}{(n + 1)^2}). Then (a_{n+1}=\frac{(2(n + 1)+1)((n + 1)-1)}{((n + 1)+1)^2}=\frac{(2n+3)(n)}{(n + 2)^2}). Calculate the ratio (\left|\frac{a_{n + 1}}{a_n}\right|=\left|\frac{\frac{(2n + 3)n}{(n + 2)^2}}{\frac{(2n+1)(n - 1)}{(n + 1)^2}}\right|=\left|\frac{(2n + 3)n(n + 1)^2}{(2n+1)(n - 1)(n + 2)^2}\right|).

Step2: Expand the polynomials

Expand the numerator and denominator: ((2n + 3)n(n + 1)^2=(2n + 3)n(n^2+2n + 1)=(2n^2+3n)(n^2+2n + 1)=2n^4 + 4n^3+2n^2+3n^3+6n^2+3n=2n^4+7n^3 + 8n^2+3n). ((2n+1)(n - 1)(n + 2)^2=(2n^2 - n - 1)(n^2+4n + 4)=2n^4+8n^3+8n^2 - n^3-4n^2-4n - n^2-4n - 4=2n^4+7n^3+3n^2-8n - 4).

Step3: Find the limit as (n\to\infty)

(\lim_{n\to\infty}\left|\frac{a_{n + 1}}{a_n}\right|=\lim_{n\to\infty}\frac{2n^4+7n^3 + 8n^2+3n}{2n^4+7n^3+3n^2-8n - 4}). Divide both the numerator and denominator by (n^4): (\lim_{n\to\infty}\frac{2+\frac{7}{n}+\frac{8}{n^2}+\frac{3}{n^3}}{2+\frac{7}{n}+\frac{3}{n^2}-\frac{8}{n^3}-\frac{4}{n^4}} = 1). Since the ratio - test is inconclusive when (\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = 1), we use the limit - comparison test. Compare (a_n=\frac{(2n + 1)(n - 1)}{(n + 1)^2}) with (b_n = 1). (\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{(2n + 1)(n - 1)}{(n + 1)^2}=\lim_{n\to\infty}\frac{2n^2 - n - 1}{n^2+2n + 1}). Divide both numerator and denominator by (n^2): (\lim_{n\to\infty}\frac{2-\frac{1}{n}-\frac{1}{n^2}}{1+\frac{2}{n}+\frac{1}{n^2}}=2\neq0). Since (\sum_{n = 1}^{\infty}b_n=\sum_{n = 1}^{\infty}1) diverges (it is a constant - term series), by the limit - comparison test, (\sum_{n = 1}^{\infty}\frac{(2n + 1)(n - 1)}{(n + 1)^2}) diverges.

Answer:

The series diverges.