calculate Σ_(n=0) ^∞ ((n+3)/(2n+1))x^n


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Question Number 35051 by math khazana by abdo last updated on 14/May/18

$c a l c u l a t e \sum_{n = 0}^{\infty} \frac{n + 3}{2 n + 1} x^{n}$
Commented by math khazana by abdo last updated on 10/Jun/18
$case1 x>0 let S(x)=Σ_(n=0) ^∞ ((n+3)/(2n+1)) x^n = (1/(√x))Σ_(n=0) ^∞ ((n+3)/(2n+1)) ((√x))^(2n+1) =(1/(√x)) f((√x)) with f(t) = Σ_(n=0) ^∞ ((n+3)/(2n+1)) t^(2n+1) f^′ (t)= Σ_(n=0) ^∞ (n+3)t^(2n) = Σ_(n=0) ^∞ n t^(2n) +3 Σ_(n=0) ^∞ t^(2n) but Σ_(n=0) ^∞ t^(2n) = (1/(1−t^2 )) and we have Σ_(n=0) ^∞ t^n =(1/(1−t)) ⇒ Σ_(n=1) ^∞ n t^(n−1) = (t/((1−t)^2 )) ⇒ Σ_(n=0) ^∞ n t^(2n) = Σ_(n=1) ^∞ n (t^2 )^n =t^2 Σ_(n=1) ^∞ n(t^2 )^(n−1) =t^2 (t^2 /((1−t^2 )^2 )) = (t^4 /((1−t^2 )^2 )) ⇒ f^′ (t) = (t^4 /((1−t^2 )^2 )) +(3/(1−t^2 )) ⇒ f(t) = ∫_. ^t ( (x^4 /((1−x^2 )^2 )) +(3/(1−x^2 )))dx = ∫_. ^t (x^4 /((1−x^2 )^2 ))dx +3 ∫_. ^t (dx/(1−x^2 )) but ∫_. ^t (dx/(1−x^2 )) = ∫^t { (1/(1−x)) +(1/(1+x))}dx=ln∣((1+t)/(1−t))∣ +c changement x =chu give ∫^t (x^4 /((1−x^2 )^2 ))dx = ∫^t ((ch^4 u)/(sh^4 u)) .shu du =∫^t ((ch^4 u)/(sh^3 u)) du = (1/4)∫^t (((1+ch(2u))^2 )/(((ch(2u)−1)/2) sh(u))) du =(1/2) ∫^t ((1+2ch(2u) +ch^2 (2u))/((ch(2u)−1)sh(u))) du ....be continued..$
$c a s e 1 x > 0 l e t S (x) = \sum_{n = 0}^{\infty} \frac{n + 3}{2 n + 1} x^{n}$ $= \frac{1}{\sqrt{x}} \sum_{n = 0}^{\infty} \frac{n + 3}{2 n + 1} {(\sqrt{x})}^{2 n + 1} = \frac{1}{\sqrt{x}} f (\sqrt{x}) w i t h$ $f (t) = \sum_{n = 0}^{\infty} \frac{n + 3}{2 n + 1} t^{2 n + 1}$ $f^{^{'}} (t) = \sum_{n = 0}^{\infty} (n + 3) t^{2 n}$ $= \sum_{n = 0}^{\infty} n t^{2 n} + 3 \sum_{n = 0}^{\infty} t^{2 n} b u t$ $\sum_{n = 0}^{\infty} t^{2 n} = \frac{1}{1 - t^{2}} a n d w e h a v e \sum_{n = 0}^{\infty} t^{n} = \frac{1}{1 - t} \Rightarrow$ $\sum_{n = 1}^{\infty} n t^{n - 1} = \frac{t}{{(1 - t)}^{2}} \Rightarrow$ $\sum_{n = 0}^{\infty} n t^{2 n} = \sum_{n = 1}^{\infty} n {(t^{2})}^{n} = t^{2} \sum_{n = 1}^{\infty} n {(t^{2})}^{n - 1}$ $= t^{2} \frac{t^{2}}{{(1 - t^{2})}^{2}} = \frac{t^{4}}{{(1 - t^{2})}^{2}} \Rightarrow$ $f^{^{'}} (t) = \frac{t^{4}}{{(1 - t^{2})}^{2}} + \frac{3}{1 - t^{2}} \Rightarrow$ $f (t) = \int_{.}^{t} (\frac{x^{4}}{{(1 - x^{2})}^{2}} + \frac{3}{1 - x^{2}}) d x$ $= \int_{.}^{t} \frac{x^{4}}{{(1 - x^{2})}^{2}} d x + 3 \int_{.}^{t} \frac{d x}{1 - x^{2}} b u t$ $\int_{.}^{t} \frac{d x}{1 - x^{2}} = \int^{t} {\frac{1}{1 - x} + \frac{1}{1 + x}} d x = l n ∣ \frac{1 + t}{1 - t} ∣ + c$ $c h a n g e m e n t x = c h u g i v e$ $\int^{t} \frac{x^{4}}{{(1 - x^{2})}^{2}} d x = \int^{t} \frac{{c h}^{4} u}{{s h}^{4} u} . s h u d u$ $= \int^{t} \frac{{c h}^{4} u}{{s h}^{3} u} d u = \frac{1}{4} \int^{t} \frac{{(1 + c h (2 u))}^{2}}{\frac{c h (2 u) - 1}{2} s h (u)} d u$ $= \frac{1}{2} \int^{t} \frac{1 + 2 c h (2 u) + {c h}^{2} (2 u)}{(c h (2 u) - 1) s h (u)} d u . . . . b e c o n t i n u e d . .$
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