How do you find a power series converging to $f(x)=x^2e^x$ and determine the radius of convergence?

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Answer 1 · 2017-01-16T08:46:04

$x^2+x^3/(1!)+x^4/(2!)+...+x^(n+2)/(n!)+..., x in (-oo, oo)$ . See the graph, with local maximum $4/e^2=0.54$ , nearly $, at$ x = -1/2#.

$1+x+x^2/(2!)+...+x^n/(n!)+... to e^x, x in (-oo, oo)$

So converges (a polynomial) $xx$ (the above series),

for $x in (-oo, oo)$ . And so,

$x^2+x^3/(1!)+x^4/(2!)+...+x^(n+2)/(n!)+...$

converges to $x^2e^x, x in (-oo, oo)$ .

graph{y(y-x^2e^x)(x+2+.001y)=0 [-10, 10, -5, 5]}

How do you find a power series converging to f(x)=x^2e^xf(x)=x^2e^x and determine the radius of convergence?