find-the-sum-of-n-0-n-2-3n-1-e-n-

Question Number 73231 by mathmax by abdo last updated on 08/Nov/19

f i n d t h e s u m o f \sum_{n = 0}^{\infty} (n^{2} - 3 n + 1) e^{- n}

Commented by mathmax by abdo last updated on 09/Nov/19

w e h a v e \sum_{n = 0}^{\infty} (n^{2} - 3 n + 1) e^{- n} = \sum_{n = 0}^{\infty} n^{2} {(e^{- 1})}^{n} - 3 \sum_{n = 0}^{\infty} n {(e^{- 1})}^{n}

+ \sum_{n = 0}^{\infty} e^{- n} f i r s t \sum_{n = 0}^{\infty} e^{- n} = \sum_{n = 0}^{\infty} {(e^{- 1})}^{n} = \frac{1}{1 - e^{- 1}} = \frac{e}{e - 1}

l e t c a l c u l a t e f (x) = \sum_{n = 0}^{\infty} {n x}^{n} a n d g (x) = \sum_{n = 0}^{\infty} n^{2} x^{n} w i t h ∣ x ∣< 1

w e h a v e \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow \sum_{n = 1}^{\infty} {n x}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow

\sum_{n = 1}^{\infty} n x^{n} = \frac{x}{{(1 - x)}^{2}} \Rightarrow \sum_{n = 0}^{\infty} {n e}^{- n} = \frac{e^{- 1}}{{(1 - e^{- 1})}^{2}} = \frac{e^{2}}{{(e - 1)}^{2}} a l s o

\sum_{n = 1}^{\infty} n^{2} x^{n - 1} = \frac{{(1 - x)}^{2} - x (- 2) (1 - x)}{{(1 - x)}^{4}} = \frac{1 - x + 2 x}{{(1 - x)}^{3}} = \frac{x + 1}{{(1 - x)}^{3}} \Rightarrow

\sum_{n = 0}^{\infty} n^{2} x^{n} = \frac{x^{2} + x}{{(1 - x)}^{3}} \Rightarrow \sum_{n = 0}^{\infty} n^{2} e^{- n} = \frac{e^{- 2} + e^{- 1}}{{(1 - e^{- 1})}^{3}}

= \frac{e + e^{2}}{{(e - 1)}^{3}} \Rightarrow \sum_{n = 0}^{\infty} (n^{2} - 3 n + 1) e^{- n} = \frac{e + e^{2}}{{(e - 1)}^{3}} - \frac{3 e^{2}}{{(e - 1)}^{2}} + \frac{e}{e - 1}

= \frac{e + e^{2} - 3 e^{2} (e - 1) + e {(e - 1)}^{2}}{{(e - 1)}^{3}} = \frac{e + e^{2} - 3 e^{3} + 3 e^{2} + e (e^{2} - 2 e + 1)}{{(e - 1)}^{3}}

= \frac{- 3 e^{3} + 4 e^{2} + e + e^{3} - 2 e^{2} + e}{{(e - 1)}^{3}} = \frac{- 2 e^{3} + 2 e^{2} + 2 e}{{(e - 1)}^{3}}

Answered by mind is power last updated on 08/Nov/19

n^{2} - 3 n + 1 = (n + 1) (n + 2) - 6 (n + 1) + 5

\sum_{n ⩾ 0} (n + 1) (n + 2) e^{- n} - 6 \sum_{n ⩾ 0} (n + 1) e^{- n} + 5 \sum_{n ⩾ 0} e^{- n}

\frac{1}{1 - x} = Σ x^{k} \Rightarrow \frac{1}{{(1 - x)}^{2}} . \sum_{k ⩾ 1} {kx}^{k - 1} = \sum_{k ⩾ 0} (k + 1) x^{k}

x = e^{- 1} \Rightarrow Σ e^{- n} = \frac{1}{1 - e^{- 1}} = \frac{e}{e - 1}

\Rightarrow x = e^{- 1} \Rightarrow \frac{1}{{(1 - e^{- 1})}^{2}} = \sum_{n ⩾ 0} (n + 1) e^{- n} = \frac{e^{2}}{(e - 1)}

\frac{1}{{(1 - x)}^{2}} = \sum_{k ⩾ 0} (k + 1) x^{k} \Rightarrow \sum_{k ⩾ 1} k (k + 1) x^{k - 1} = \frac{2}{{(1 - x)}^{3}} = \sum_{k ⩾ 0} (k + 1) (k + 2) x^{k}

x = e^{- 1} \Rightarrow \sum_{n ⩾ 0} (n + 1) (n + 2) e^{- n} = \frac{2}{{(1 - e^{- 1})}^{3}} = \frac{{2 e}^{3}}{{(e - 1)}^{3}}

so we get \frac{{2 e}^{3}}{{(e - 1)}^{3}} - \frac{{6 e}^{2}}{{(e - 1)}^{2}} + \frac{5 e}{(e - 1)}

= \frac{e^{3} - {4 e}^{2} + 5 e}{{(e - 1)}^{3}}

Commented by mathmax by abdo last updated on 08/Nov/19

t h a n k x s i r .