Sum of the Infinite Power Series

Question

I need to calculate the sum of the infinite power series

$$\sum_{k=0}^\infty\frac{2^k(k+1)k}{3e^2k!}$$

I was thinking of using the exponential function power series expansion formula

$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$

but still not sure how to deal with the initial expression to make use of the formula.

Answer 1

hint

For $ k\ge 2 $, Replace $(k+1)k $ by

$$k(k-1)+2k$$

to get $$\frac{2^kk(k+1)}{k!}=4\Bigl(\frac{2^{k-2}}{(k-2)!}+\frac{2^{k-1}}{(k-1)!}\Bigr)$$

Answer 2

Hint: What is the MacLaurin series of $x^2 \exp(x)$?

Answer 3

We have

\begin{align}\sum_{k=0}^\infty\frac{2^k(k+1)k}{3e^2k!}&=\frac{1}{3e^2}\sum_{k=0}^\infty\frac{2^k(k+1)k}{k!}\\&=\frac{1}{3e^2}\sum_{k=0}^\infty\frac{2^k(k^2+k)}{k!}\\&=\frac{1}{3e^2}\sum_{k=0}^\infty\frac{2^kk^2}{k!}+\frac{1}{3e^2}\sum_{k=0}^\infty\frac{2^kk}{k!}\end{align}

where the Maclaurin expansion of $e^2$ is

$$e^2=\sum_{k=0}^{\infty}\frac{2^k}{k!}$$

therefore

$$\sum_{k=0}^{\infty}\frac{2^kk^2}{k!}=6\sum_{k=0}^{\infty}\frac{2^k}{k!}=6e^2$$ $$\sum_{k=0}^{\infty}\frac{2^kk}{k!}=2\sum_{k=0}^{\infty}\frac{2^k}{k!}=2e^2$$

thus

\begin{align}\sum_{k=0}^\infty\frac{2^k(k+1)k}{3e^2k!}&=\frac{1}{3e^2}\sum_{k=0}^\infty\frac{2^kk^2}{k!}+\frac{1}{3e^2}\sum_{k=0}^\infty\frac{2^kk}{k!}\\&= \frac{1}{3e^2}6e^2+\frac{1}{3e^2}2e^2\\&= 2+\frac23\\&= \frac{8}{3} \end{align}