Manipulating infinite summation I want to solve for the $$\sum_{b=0}^{\infty} \binom{a+b}{b}e^{0.3b}(0.3)^{a+b} $$
Can anyone give me idea how to start this? I'm thinking of manipulating it to look like the binomial series $$\sum_{j=0}^{n} \binom{n}{j}a^{j}b^{n-j} $$ but I have no idea regarding the changing of their upper limits. Any help would be greatly appreciated ☺
 A: You can write:
$$\sum_{b=0}^{\infty}\binom{a+b}{b}\exp(0.3 b)0.3^{a+b} = (0.3)^a\sum_{b=0}^{\infty}\binom{a+b}{b}x^b $$
where $x = 0.3\exp(0.3)$
We can then compute this summation by noting that:
$$\binom{a+b}{b} = \frac{1}{a!}(b+a)(b+a-1)\cdots(b+1)$$
Then to compute the desired summation we can take the geometric series
$$\sum_{b=0}^{\infty}x^{a+b}= \frac{x^a}{1-x}$$
and differentiate both sides $a$ times.
A: We can apply the binomial series expansion
\begin{align*}
\sum_{b=0}^\infty\binom{\alpha}{b}x^b=(1+x)^\alpha\qquad\qquad |x|<1, \alpha\in\mathbb{C}\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{b=0}^\infty\binom{a+b}{b}e^{0.3b}(0.3)^{a+b}}
&=0.3^a\sum_{b=0}^\infty\binom{-a-1}{b}\left(-0.3e^{0.3}\right)^b\tag{2}\\
&\color{blue}{=\frac{0.3^a}{\left(1-0.3e^{0.3}\right)^{a+1}}}\tag{3}
\end{align*}

Comment:


*

*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (3) we apply the binomial series expansion according to (1) noting that $|-0.3e^{0.3}|<1$.
