calculate-n-0-2n-3-n-1-2-n-2-2-

Question Number 32997 by abdo imad last updated on 09/Apr/18

c a l c u l a t e \sum_{n = 0}^{\infty} \frac{2 n + 3}{{(n + 1)}^{2} {(n + 2)}^{2}}

Commented by abdo imad last updated on 10/Apr/18

l e t f i n d a a n d b / F (n) = \frac{2 n + 3}{{(n + 1)}^{2} {(n + 2^{})}^{2}} = \frac{a}{{(n + 1)}^{2}} + \frac{b}{{(n + 2)}^{2}}

w e h a v e {l i m}_{n \to + \infty} {(n + 1)}^{2} F (n) = 0 = a + b \Rightarrow b = - a

F (0) = \frac{3}{4} = a + \frac{b}{4} \Rightarrow 3 = 4 a + b = a + b + 3 a = 3 a \Rightarrow a = 1

a n d b = - 1 \Rightarrow \frac{2 n + 3}{{(n + 1)}^{2} {(n + 2)}^{2}} = \frac{1}{{(n + 1)}^{2}} - \frac{1}{{(n + 2)}^{2}}

l e t p u t S_{N} = \sum_{n = 0}^{N} \frac{2 n + 3}{{(n + 1)}^{2} {(n + 2)}^{2}} a n d

S = \sum_{n = 0}^{\infty} \frac{2 n + 3}{{(n + 1)}^{2} {(n + 2)}^{2}} w e h a v e S = {l i m}_{N \to \infty} S_{N}

b u t S_{N} = \sum_{n = 0}^{N} (\frac{1}{{(n + 1)}^{2}} - \frac{1}{{(n + 2)}^{2}}) = \sum_{n = 0}^{N} (u_{n} - u_{n + 1})

w i t h u_{n} = \frac{1}{{(n + 1)}^{2}} \Rightarrow S_{N} = u_{0} - u_{1} + u_{1} - u_{2} + \dots u_{n} - u_{n + 1}

= u_{0} - u_{n + 1} = 1 - \frac{1}{{(n + 2)}^{2}} \Rightarrow {l i m}_{n \to + \infty} S_{N} = 1

\Rightarrow S = 1 .