Computing:$\sum_{n=0}^\infty\frac{3^n}{n!(n+3)}$

Question

I'm learning the subject Power Series and I can't figure out how to find the sum of the series $$\sum_{n=0}^\infty\frac{3^n}{n!(n+3)}$$.

I know that the power series $\sum_{n=0}^\infty\frac{3^n}{n!(n+3)}x^n$ is convergence for every $x$, and uniform convergence for every $[-r, r], r>0$ because $\lim\limits_{n \to\infty}|\frac{a_n}{a_{n+1}}|=\infty$. I don't know how to proceed, I think I should use the information that this power series convergence uniformly and then somehow Integrating/differentiating this Power Series.

Thanks!

Answer 1

$$ \sum_{n=0}^{+\infty}\frac{3^n}{n!\left(n+3\right)}=\frac{1}{27}\sum_{n=0}^{+\infty}\frac{1}{n!}\frac{3^{n+3}}{n+3} $$ And $$ \frac{3^{n+3}}{n+3}=\int_{0}^{3}x^{n+2}\text{d}x $$ Hence it becomes ( using uniform convergence on every compact of $\mathbb{R}$ especially $\left[0,3\right]$ ) $$ \sum_{n=0}^{+\infty}\frac{3^n}{n!\left(n+3\right)}=\frac{1}{27}\sum_{n=0}^{+\infty}\frac{1}{n!}\int_{0}^{3}x^{n+2}\text{d}x=\frac{1}{27}\int_{0}^{3}\sum_{n=0}^{+\infty}\frac{1}{n!}x^{n+2}\text{d}x $$ Then $$\sum_{n=0}^{+\infty}\frac{3^n}{n!\left(n+3\right)}=\frac{1}{27}\int_{0}^{3}x^2e^{x}\text{d}x$$ Hence you conlude ( integrating by parts ) $$ \sum_{n=0}^{+\infty}\frac{3^n}{n!\left(n+3\right)}=\frac{1}{27}\left(5e^3-2\right) $$

Answer 2

You can merely use the differentiation of power series as follows

$$\sum_{n=0}^\infty\frac{3^n}{n!(n+3)}= \sum_{n=0}^\infty\frac{(n+2)(n+1)3^n}{(n+3)!}=\sum_{n=3}^\infty\frac{(n-1)(n-2)3^{n-3}}{n!}\\= \frac{d^2}{dx^2}\left( \sum_{n=1}^\infty\frac{x^{n-1}}{n!}\right)\Bigg|_{x=3}= \frac{d^2}{dx^2}\left( -\frac{1}{x}+\frac1x\sum_{n=0}^\infty\frac{x^{n}}{n!}\right)\Bigg|_{x=3}\\=\frac{d^2}{dx^2}\left(-\frac{1}{x} +\frac{e^{x}}{x}\right)\Bigg|_{x=3}= \left(-\frac{2}{x^3} +\frac{e^{x}(x^2-2x+2)}{x^3}\right)\Bigg|_{x=3} \\ = \color{red}{\frac{1}{27}\left(5e^3-2\right)}$$

Answer 3

The approach with hypergeometric functions is $$ \sum_{n\ge 0}\frac{1}{n+3} \frac{3^n}{n!} = \sum_{n\ge 0}\frac{(3)_n}{3(4)_n} \frac{3^n}{n!} = \frac{1}{3}{}_1F_1(3;4;3). $$

Kummer's transformation yields $$ _1F_1(3;4;3) = e^3{}_1F_1(1;4,-3). $$

A contiguous relation of this is (see for example the Abramowitz-Stegun book formula (13.4.6), at $a=1, b=4, z=-3$) $$ -3{}_1F_1(1;4;-3)+3{}_1F_1(0;4;-3) + (-3)_1F_1(1;3;-3)=0. $$ So $$ -{}_1F_1(1;4;-3)+1 -{} _1F_1(1;3;-3)=0. $$ $$ {}_1F_1(1;4;-3)= 1-{}_1F_1(1;3;-3), $$ and this is already in the Wikipedia article $$ _1F_1(1;3;-3) = 2(e^{-3}-1+3)/9 = \frac29(e^{-3}+2). $$ Backward insertion gives the result.