2
$\begingroup$

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 ☺

$\endgroup$
1
$\begingroup$

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 $|-0.3e^{0.3}|<1$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.