Find the value of: Σ_(n = 1) ^∞ ((n^2 + 1)/(n + 2)). (x^n /(n!))


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Question Number 61922 by Tawa1 last updated on 11/Jun/19

$Find the value of : \underset{n = 1}{\sum^{\infty}} \frac{n^{2} + 1}{n + 2} . \frac{x^{n}}{n!}$
Commented by maxmathsup by imad last updated on 12/Jun/19
$let p(x) =Σ_(n=1) ^∞ ((n^2 +1)/(n+2)) (x^(n+2) /(n!)) ⇒((p(x))/x^2 ) =Σ_(n=1) ^∞ ((n^2 +1)/(n+2)) (x^n /(n!)) let find p(x) p^′ (x) =Σ_(n=1) ^∞ ((n^2 +1)/(n!)) x^(n+1) =x{ Σ_(n=1) ^∞ (n^2 /(n!)) x^n +Σ_(n=1) ^∞ (x^n /(n!))} Σ_(n=1) ^∞ (x^n /(n!)) =e^x −1 Σ_(n=1) ^∞ (n^2 /(n!))x^n =Σ_(n=1) ^∞ (n/((n−1)!))x^n =Σ_(n=1) ^∞ ((n−1 +1)/((n−1)!)) x^n =Σ_(n=1) ^∞ ((n−1)/((n−1)!))x^n +Σ_(n=1) ^∞ (x^n /((n−1)!)) =Σ_(n=2) ^∞ (x^n /((n−2)!)) +Σ_(n=0) ^∞ (x^(n+1) /(n!)) =Σ_(n=0) ^∞ (x^(n+2) /(n!)) +Σ_(n=0) ^∞ (x^(n+1) /(n!)) =x^2 e^x +x e^x =(x^2 +x)e^x ⇒p^′ (x) =x{ (x^2 +x)e^x +e^x −1} =x{(x^2 +x+1)e^x −1} =(x^3 +x^2 +x)e^x −x ⇒p(x) =∫(x^3 +x^2 +x)e^x dx−(x^2 /2) +c ∫xe^x =xe^x −∫e^x dx =xe^x −e^x =(x−1)e^x ∫ x^2 e^x dx =x^2 e^x −∫2x e^x dx =x^2 e^x −2(x−1)e^x =(x^2 −2x+2)e^x ∫x^3 e^x dx =x^3 e^x −∫ 3x^2 e^x dx =x^3 e^x −3(x^2 −2x+2)e^x =(x^3 −3x^2 +6x −6)e^x ⇒ p(x) =(x^3 −3x^2 +6x−6 +x^2 −2x+2 +x−1)e^x −(x^2 /2) +c =(x^3 −2x^2 +5x −5)e^x −(x^2 /2) +c p(0) =0 =−5 +c ⇒c =5 ⇒p(x) =(x^3 −2x^2 +5x−5)e^x −(x^2 /2) +5 ⇒ Σ_(n=1) ^∞ ((n^2 +1)/(n+2)) (x^n /(n!)) =((p(x))/x^2 ) =(x−2+(5/x) −(5/x^2 ))e^x −(1/2) +(5/x^2 ) ( x≠0) .$
$l e t p (x) = \sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n + 2} \frac{x^{n + 2}}{n!} \Rightarrow \frac{p (x)}{x^{2}} = \sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n + 2} \frac{x^{n}}{n!} l e t f i n d p (x)$ $p^{^{'}} (x) = \sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n!} x^{n + 1} = x {\sum_{n = 1}^{\infty} \frac{n^{2}}{n!} x^{n} + \sum_{n = 1}^{\infty} \frac{x^{n}}{n!}}$ $\sum_{n = 1}^{\infty} \frac{x^{n}}{n!} = e^{x} - 1$ $\sum_{n = 1}^{\infty} \frac{n^{2}}{n!} x^{n} = \sum_{n = 1}^{\infty} \frac{n}{(n - 1)!} x^{n} = \sum_{n = 1}^{\infty} \frac{n - 1 + 1}{(n - 1)!} x^{n} = \sum_{n = 1}^{\infty} \frac{n - 1}{(n - 1)!} x^{n} + \sum_{n = 1}^{\infty} \frac{x^{n}}{(n - 1)!}$ $= \sum_{n = 2}^{\infty} \frac{x^{n}}{(n - 2)!} + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n!} = \sum_{n = 0}^{\infty} \frac{x^{n + 2}}{n!} + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n!} = x^{2} e^{x} + x e^{x}$ $= (x^{2} + x) e^{x} \Rightarrow p^{^{'}} (x) = x {(x^{2} + x) e^{x} + e^{x} - 1} = x {(x^{2} + x + 1) e^{x} - 1}$ $= (x^{3} + x^{2} + x) e^{x} - x \Rightarrow p (x) = \int (x^{3} + x^{2} + x) e^{x} d x - \frac{x^{2}}{2} + c$ $\int {x e}^{x} = {x e}^{x} - \int e^{x} d x = {x e}^{x} - e^{x} = (x - 1) e^{x}$ $\int x^{2} e^{x} d x = x^{2} e^{x} - \int 2 x e^{x} d x = x^{2} e^{x} - 2 (x - 1) e^{x} = (x^{2} - 2 x + 2) e^{x}$ $\int x^{3} e^{x} d x = x^{3} e^{x} - \int 3 x^{2} e^{x} d x = x^{3} e^{x} - 3 (x^{2} - 2 x + 2) e^{x}$ $= (x^{3} - 3 x^{2} + 6 x - 6) e^{x} \Rightarrow$ $p (x) = (x^{3} - 3 x^{2} + 6 x - 6 + x^{2} - 2 x + 2 + x - 1) e^{x} - \frac{x^{2}}{2} + c$ $= (x^{3} - 2 x^{2} + 5 x - 5) e^{x} - \frac{x^{2}}{2} + c$ $p (0) = 0 = - 5 + c \Rightarrow c = 5 \Rightarrow p (x) = (x^{3} - 2 x^{2} + 5 x - 5) e^{x} - \frac{x^{2}}{2} + 5 \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n + 2} \frac{x^{n}}{n!} = \frac{p (x)}{x^{2}} = (x - 2 + \frac{5}{x} - \frac{5}{x^{2}}) e^{x} - \frac{1}{2} + \frac{5}{x^{2}} (x \neq 0) .$
Commented by Tawa1 last updated on 12/Jun/19

$God bless you sir$
Commented by Mr X pcx last updated on 12/Jun/19

$y o u a r e w e l c o m e s i r .$
	Answered by Smail last updated on 12/Jun/19

	$\frac{n^{2} + 1}{n + 2} = \frac{(n + 2) (n - 2) + 5}{n + 2} = n - 2 + \frac{5}{n + 2}$ $L e t f (x) = \underset{n = 1}{\sum^{\infty}} \frac{n^{2} + 1}{n + 2} \times \frac{x^{n}}{n!}$ $f (x) = \underset{n = 1}{\sum^{\infty}} (n - 2 + \frac{5}{n + 2}) \frac{x^{n}}{n!}$ $= \underset{n = 1}{\sum^{\infty}} \frac{n}{n!} x^{n} - 2 \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n!} + 5 \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{(n + 2) n!}$ $= \underset{n = 0}{\sum^{\infty}} \frac{x^{n + 1}}{n!} - 2 (\underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} - 1) + \frac{5}{x^{2}} \underset{n = 1}{\sum^{\infty}} \frac{x^{n + 2}}{(n + 2) n!}$ $= {x e}^{x} - 2 (e^{x} - 1) + \frac{5}{x^{2}} p (x)$ $p (x) = \underset{n = 1}{\sum^{\infty}} \frac{x^{n + 2}}{(n + 2) n!}$ $p^{'} (x) = \underset{n = 1}{\sum^{\infty}} \frac{x^{n + 1}}{n!} = x (e^{x} - 1)$ $p (x) = \int_{0}^{x} ({t e}^{t} - t) d t = {[e^{t} (t - 1) - \frac{t^{2}}{2}]}_{0}^{x}$ $p (x) = e^{x} (x - 1) - \frac{x^{2}}{2} + 1$ $f (x) = {x e}^{x} - 2 e^{x} + 2 + \frac{5}{x^{2}} (e^{x} (x - 1) - \frac{x^{2}}{2} + 1)$ $f (x) = \underset{n = 1}{\sum^{\infty}} \frac{n^{2} + 1}{(n + 2) n!} x^{n} = e^{x} (x - 2 + \frac{5 (x - 1)}{x^{2}}) + \frac{5}{x^{2}} - \frac{1}{2}$
	Commented by Tawa1 last updated on 12/Jun/19

	$God bless you sir$
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