I need to find closed form expressions for these power series F and G:
$a_n = 19^n + n(-5)^n$ for n .. elements of natural numbers including 0
$F(x) = \sum_{n=0}^\infty a_nx^n$
and
$G(x) = \sum_{n=0}^\infty \frac{a_n}{n!}x^n$
Regarding F(x): i tried to split it up in two series and find a solution through geometric series - iam not sure if my approach is correct:
$19^n x^n = (19x)^n$ and $x^nn(-5)^n = n(-5x)^n$
$F(x) = \sum_{n=0}^\infty (19x)^n + \sum_{n=0}^\infty n(-5x)^n$
$F(x) = \frac{1}{1-19x} + \sum_{n=0}^\infty n(-5x)^n$
what do i do with $\sum_{n=0}^\infty n(-5x)^n$?
$F(x) = \frac{1}{1-19x} - \frac{5}{(1+5x)^2}$ - is this correct? Really not sure about second part
Regarding G(x):
$G(x) = \sum_{n=0}^\infty \frac{a_n}{n!}x^n$
I am well aware of $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ but how do i deal with $19^n$ or $n(-5)^n$ in the following:
$G(x) = \sum_{n=0}^\infty \frac{19^n}{n!}x^n + \sum_{n=0}^\infty \frac{n(-5)^n}{n!}x^n$
I know this is a lot of questions, but i do hope that someone can help me out. Thanks in advance!!!
I will give here solution to the last example: $$ \sum_{n=0}^\infty n\frac{(ax)^n}{n!}=\sum_{n=1}^\infty\frac{(ax)^n}{(n-1)!} =ax \sum_{n=1}^\infty\frac{(ax)^{n-1}}{(n-1)!}\stackrel{k=n-1}= ax \sum_{k=0}^\infty\frac{(ax)^k}{k!}=ax e^{ax}. $$