Can you find a series whose sum is $.... = (a) *\sqrt{\frac{1}{1-n}}$ I discovered here that $ \frac{1}{n-1}= (n-1)^-1$ is the sum om a geometric series:
$$\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{1}{n-1}$$
can you tell me if $(\sqrt{1-n})^-1$ $$\sqrt{\frac{1}{1-n}}$$
can be considered the result/sum of any kind of series, sequence, progression 
Edit:
consider $|n| <1$,  specify what happens if we have a factor $a$ before the square root and, if you please,  case consider the particular case when the sum of the series is  $$0.7071 * \sqrt{\frac{1}{1-.99}}$$

$\sum_{n=m}^{\infty} (a_{n}-a_{n+1})
=a_m
$,
so
$\begin{array}\\
\frac1{\sqrt{m}}
&=\sum_{n=m}^{\infty} (\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}})\\
&=\sum_{n=m}^{\infty} (\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}})\frac{\frac1{\sqrt{n}}+\frac1{\sqrt{n+1}}}{\frac1{\sqrt{n}}+\frac1{\sqrt{n+1}}}\\
&=\sum_{n=m}^{\infty} \frac{\frac1{n}-\frac1{n+1}}{\frac1{\sqrt{n}}+\frac1{\sqrt{n+1}}}\\
&=\sum_{n=m}^{\infty} \frac{1}{n(n+1)(\frac1{\sqrt{n}}+\frac1{\sqrt{n+1}})}\\
&=\sum_{n=m}^{\infty} \frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}\\
\end{array}
$
