I want to calculate
$$\sum_{k=0}^\infty\binom{k+3}k(0.2)^k$$
to get the exact value of it. I have excel and other tools to help me so it is fine if it is computationally expensive. Is there a clear and repeatable way to solve this infinite series? Thank you. This is my first post and be sure to give me some suggestions as well.
It’s a general fact that
$$\sum_{k\ge 0}\binom{k+n}nx^k=\frac1{(1-x)^{n+1}}\;.$$
You can prove this by induction on $n$, starting with the geometric series
$$\frac1{1-x}=\sum_{k\ge 0}x^k$$
and differentiating repeatedly with respect to $x$. You want the case $n=3$:
$$\sum_{k\ge 0}\binom{k+3}kx^k=\sum_{k\ge 0}\binom{k+3}3x^k=\frac1{(1-x)^4}\,.$$
Now just substitute $x=0.2$.
Using the negative binomial coefficient $\binom{-4}{k}$, we have $$ \begin{align} \sum_{k=0}^\infty\binom{k+3}{k}(0.2)^k &=\sum_{k=0}^\infty\binom{-4}k(-1)^k(0.2)^k\\ &=(1-0.2)^{-4}\\[6pt] &=\frac{625}{256} \end{align} $$
