What is the sum of this series: $1 + \frac{1}{5}x + \frac{1 \times 6}{5 \times 10}x^2 +\cdots$?

Question

Say I have a series like the following;

$$1 + \frac{1}{5}x + \frac{1 \times 6}{5 \times 10}x^2 + \frac{1 \times 6 \times 11}{5 \times 10 \times 15}x^3 + \cdots.$$

How do I find the sum of this? I'm trying to find an expression for the nth term but struggling!

Answer 1

As @Jennifer answered, the general term seems to be $$ \frac{\prod_{i=0}^{n-1} (5i+1) }{\prod_{i=0}^{n-1} 5(i+1)}=\frac{\Gamma\left(\frac15+n\right)}{\Gamma\left(\frac15\right)\Gamma\left(n+1\right)}=\frac{\left(\frac15\right)_n}{n!} \tag1 $$ where $(x)_n=x(x+1)\cdots (x+n-1)$, then one may recall the generalized binomial theorem, $$ \left(1-x \right)^{-\alpha}=\sum_{n=0}^\infty\frac{\left(\alpha\right)_n}{n!}x^n. \tag2$$ Hence, by putting $\alpha=\dfrac15$ in $(2)$, one gets, using $(1)$:

$$ 1 + \frac{1}{5}x + \frac{1 \times 6}{5 \times 10}x^2 + \frac{1 \times 6 \times 11}{5 \times 10 \times 15}x^3 + \cdots =\color{red}{\frac1{(1-x)^{1/5}}}, \quad |x|<1. \tag3 $$

Answer 2

The term for $x^n$ seems to be : $$\frac{\prod_{i=0}^{n-1} (5i+1) }{\prod_{i=0}^{n-1} 5(i+1)} $$