# Is there a way to find the sum of an infinite series (not geometric)

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 $$x=0.2$$.

• Thanks :) It's very helpful that you gave me a general answer as well. – CELLSecret Jan 17 at 19:39
• @CELLSecret: You’re welcome. – Brian M. Scott Jan 17 at 19:40
• Is there a way to to write out the expression on stackexchange without embedding an image? If so, how? Thanks – CELLSecret Jan 17 at 19:54
• @CELLSecret Here is the typesetting tutorial : math.meta.stackexchange.com/questions/5020/… – cosmo5 Jan 17 at 19:56

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}