$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$ How to find the sum of the following series:
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$
Any hints.

I am applying Robert Israel's suggestion -
I have learned that this is generally a good thing to do.
The general term in
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot$
seems to be
$\begin{array}\\
a_n
&=(-1)^n\dfrac{\prod_{k=1}^n (3k-2)}{5^n n!}\\
&=\dfrac{\prod_{k=1}^n (2-3k)}{5^n n!}\\
&=\dfrac{3^n\prod_{k=1}^n (\frac23-k)}{5^n n!}\\
&=\dfrac{3^n\prod_{k=0}^{n-1} (\frac23-(k+1))}{5^n n!}\\
&=\dfrac{3^n\prod_{k=0}^{n-1} (-\frac13-k)}{5^n n!}\\
&=\left(\dfrac{3}{5}\right)^n\dfrac{\prod_{k=0}^{n-1} (-\frac13-k)}{ n!}\\
\end{array}
$
The expansion of
$(1+t)^r$
is
$\begin{array}\\
(1+t)^r
&=\sum_{n=0}^{\infty} t^n \dfrac{\prod_{k=0}^{n-1} (r-k)}{n!}\\
\end{array}
$
If $t=\frac35$
and
$r=-\frac13$,
this gives
$(1+\frac35)^{-1/3}
=\sum_{n=0}^{\infty} (\frac35)^n \dfrac{\prod_{k=0}^{n-1} (-\frac13-k)}{n!}
=1-\sum_{n=1}^{\infty} a_n
$.
Therefore
$\sum_{n=1}^{\infty} a_n
=1-(1+\frac35)^{-1/3}
=1-(\frac85)^{-1/3}
=1-(\frac58)^{1/3}
=1-\frac{5^{1/3}}{2}
$.
Cowabunga!!!!!
Generalizations:
If $a, b, c, d$ are non-zero integers
such that
$gcd(a, b)
=gcd(c, d)
=1
$,
$\begin{array}\\
(1+\frac{a}{b})^{c/d}
&=\sum_{n=0}^{\infty} \left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (\frac{c}{d}-k)}{n!}\\
&=\sum_{n=0}^{\infty} \left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (c-kd)}{d^n n!}\\
&=\sum_{n=0}^{\infty} \dfrac{a^n\prod_{k=0}^{n-1} (c-kd)}{(bd)^n n!}\\
\end{array}
$
Putting
$-c/d$ for $c/d$,
$\begin{array}\\
(1+\frac{a}{b})^{-c/d}
&=\sum_{n=0}^{\infty} \left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (-\frac{c}{d}-k)}{n!}\\
&=\sum_{n=0}^{\infty} (-1)^n\left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (c+kd)}{d^n n!}\\
&=\sum_{n=0}^{\infty} (-1)^n\dfrac{a^n\prod_{k=0}^{n-1} (c+kd)}{(bd)^n n!}\\
\end{array}
$
