The sum of the following infinite series $\frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$ 
The sum of the following infinite series $\displaystyle \frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$

$\bf{My\; Try::}$ We can write the given series as $$\left(1+\frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots\right)-1$$
Now camparing with $$(1+x)^n = 1+nx+\frac{n(n-1)x^2}{2!}+\cdots$$
So we get $\displaystyle nx=\frac{4}{20}$ and $\displaystyle \frac{n(n-1)x^2}{2}=\frac{4\cdot 7}{20\cdot 30}$
So we get $$\frac{nx\cdot (nx-x)}{2}=\frac{4\cdot 7}{20\cdot 30}\Rightarrow \frac{4}{20}\cdot \left(\frac{4-20}{20}\right)\cdot \frac{1}{2}x^2=\frac{4}{20}\cdot \frac{7}{30}$$ 
But here $x^2=\text{Negative.}$
I did not understand how can I solve it
Help me, Thanks
 A: The numerators suggest that you could make use of a power series involving exponents that are rational numbers with denominator $3$.
$$\begin{align}
\sum_{n=1}^{\infty}\frac{4\cdot7\cdot\cdots\cdot(3n+1)}{(n+1)!10^n}
&=\sum_{n=1}^{\infty}\frac{\frac43\cdot\frac73\cdot\cdots\cdot\frac{3n+1}3}{(n+1)!\left(10/3\right)^n}\\
&=\sum_{n=1}^{\infty}\frac{1}{n+1}\binom{\frac{3n+1}{3}}{n}\left(\frac{3}{10}\right)^n\\
&=\left[\sum_{n=1}^{\infty}\frac{1}{n+1}\binom{\frac{3n+1}{3}}{n}x^n\right]_{x=3/10}\\
&=\left[\frac{1}{x}\sum_{n=1}^{\infty}\frac{1}{n+1}\binom{\frac{3n+1}{3}}{n}x^{n+1}\right]_{x=3/10}\\
&=\left[\frac{1}{x}\int_0^x\sum_{n=1}^{\infty}\binom{\frac{3n+1}{3}}{n}t^{n}\,dt\right]_{x=3/10}\\
&=\left[\frac{1}{x}\int_0^x\sum_{n=1}^{\infty}\binom{-\frac{4}{3}}{n}(-t)^{n}\,dt\right]_{x=3/10}\\
&=\left[\frac{1}{x}\int_0^x\left(\left(1-t\right)^{-4/3}-1\right)\,dt\right]_{x=3/10}\\
&=\left[\frac{1}{x}\left[3\left(1-t\right)^{-1/3}-t\right]_{t=0}^{t=x}\right]_{x=3/10}\\
&=\left[\frac{1}{x}\left(3\left(1-x\right)^{-1/3}-x-3\right)\right]_{x=3/10}\\
&=\frac{10}{3}\left(3\left(1-\frac{3}{10}\right)^{-1/3}-\frac{3}{10}-3\right)\\
&=10\left(\frac{7}{10}\right)^{-1/3}-11\\
&=\sqrt[3]{\frac{10^4}{7}}-11\\
\end{align}$$
A: Since the question asks about $X=\frac4{20}(1+\frac7{30}(1+...)))$, consider
$$1+\frac1{10}(1+\frac4{20}(1+\frac7{30}(...)))$$
The $10,20,30$ have a factor $1,2,3$ which will become $n!$ in the denominator.
Then $\frac1{10},\frac4{10},\frac7{10}$ increase by $3/10$ each time.   Take the factor $3/10$ out, and we have $\frac13,\frac43,\frac73$  which increase by 1 each time.  Let $x=3/10,n=1/3$.
$$1+xn+\frac{x^2}{2!}n(n+1)+\frac{x^3}{3!}n(n+1)(n+2)+...\\=(1-x)^{-n}=0.7^{-1/3}$$
This equals $1+\frac1{10}(1+X)$, 
so your sum is $X=10(0.7)^{-1/3}-11$
