Sum of a Hyper-geometric series. (NBHM 2011) 
How to find the sum of the following series
$$\frac{1}{5} - \frac{1\cdot 4}{5\cdot 10} + \frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15} - \dots\,.?$$

I have no idea. I have written the general term and tested its convergence by Gauss' test for convergence, but they are neither the question nor the answer.

From the Generalized Binomial Theorem (for $|x|<1$), $$\left(1+x\right)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$
Method $1:$
$\displaystyle -\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}=\frac{-\frac13\left(-\frac13-1\right)\left(-\frac13-2\right)3^3}{3!\cdot 5^3}=\frac{-\frac13\left(-\frac13-1\right)\left(-\frac13-2\right)}{3!}\left(\frac35\right)^3 $
and $\displaystyle \frac{1\cdot 4}{5\cdot 10}=\frac{-\frac13\left(-\frac13-1\right)3^2}{2!\cdot 5^2}=\frac{-\frac13\left(-\frac13-1\right)}{2!}\left(\frac35\right)^2$
and $\displaystyle-\frac15=\left(-\frac13\right)\left(\frac35\right)$
So, the given series $\displaystyle=1-\left(1+\frac35\right)^{-\frac13}$

Method $2:$
If $\displaystyle nx=-\frac15   \  \  \   \ (1)$
and $\displaystyle\frac{n(n-1)}{2!}x^2=-\frac{1\cdot4}{5\cdot10}=\frac2{25} \  \  \   \ (2)$
Divide $(2)$ with the square of $(1)$ to get $n$ and then $x$ using $(1)$
and check that $x,n$ satisfies $\displaystyle\frac{n(n-1)(n-2)}{3!}x^3=-\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}$
