calculate-n-1-n-4-2n-2-3-x-n-n-in-case-of-convergence-

Question Number 73261 by mathmax by abdo last updated on 09/Nov/19

c a l c u l a t e \sum_{n = 1}^{\infty} (n^{4} + 2 n^{2} - 3) \frac{x^{n}}{n!} i n c a s e o f c o n v e r g e n c e .

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

w e h a v e \sum_{n = 1}^{\infty} (n^{4} + 2 n^{2} - 3) \frac{x^{n}}{n!} = \sum_{n = 1}^{\infty} \frac{n^{3}}{(n - 1)!} x^{n} + 2 \sum_{n = 1}^{\infty} \frac{{n x}^{n}}{(n - 1)!}

- 3 \sum_{n = 1}^{\infty} \frac{x^{n}}{n!} w e h a v e f i r s t \sum_{n = 1}^{\infty} \frac{x^{n}}{n!} = e^{x} - 1

\sum_{n = 1}^{\infty} \frac{{n x}^{n}}{(n - 1)!} = \sum_{n = 0}^{\infty} \frac{(n + 1) x^{n + 1}}{n!} = \sum_{n = 1}^{\infty} \frac{x^{n + 1}}{(n - 1)!} + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n!}

= x \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n!} + x e^{x} = x^{2} e^{x} + {x e}^{x}

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

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

+ \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{(n - 1)!} + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n!}

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

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

+ x^{2} e^{x}

= \sum_{n = 0}^{\infty} \frac{n + 1}{n!} x^{n + 2} + 2 \sum_{n = 0}^{\infty} \frac{x^{n + 3}}{n!} + x^{2} e^{x}

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

= \sum_{n = 0}^{\infty} \frac{x^{n + 3}}{n!} + 2 x^{2} e^{x} + 2 x^{3} e^{x}

= x^{3} e^{x} + 2 x^{3} e^{x} + 2 x^{2} e^{x} = (3 x^{3} + 2 x^{2}) e^{x} \dots b e c o n t i n u e d \dots

Answered by mind is power last updated on 09/Nov/19

use 1, x, x (x - 1), x (x - 1) (x - 2), x (x - 1) (x - 2) (x - 3)

as Base of {IR}^{4} [X]

n^{4} + {2 n}^{2} - 3 = an (n - 1) (n - 2) (n - 3) + bn (n - 1) (n - 2) + cn (n - 1) + dn + t

\sum_{n ⩾ 1} (n^{4} + {2 n}^{2} - 3) \frac{x^{n}}{n!} = \sum_{n ⩾ 4} (a \frac{x^{n}}{(n - 4)!}) + b \sum_{n ⩾ 4} \frac{x^{n}}{(n - 3))!} + c \sum_{n ⩾ 4} \frac{x^{n}}{(n - 2)!} + d \sum_{n ⩾ 4} \frac{x^{n}}{(n - 1)!} + t \sum_{n ⩾ 4} \frac{x^{n}}{n!}

+ {\underset{n = 1}{\sum^{3}} (n^{4} + {2 n}^{2} - 3) \frac{x^{n}}{n!} = h (x)}

= {ax}^{4} e^{x} + {bx}^{3} (e^{x} - 1) + {cx}^{2} (e^{x} - 1 - x) + dx (e^{x} - 1 - x - \frac{x^{2}}{2}) + t (e^{x} - 1 - x - \frac{x^{2}}{2} - \frac{x^{3}}{6}) + h (x)