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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 $$0.7071 * \sqrt{\frac{1}{1-.99}}$$

Community
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    $\sum\limits_{n=1}^{\infty}\frac{a}{2^n}=a$. – Asinomás Mar 27 '16 at 06:10
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    The notation is a little dangerous: observe that what you discovered in that link is true only if $;|n|<1;$ , and since $;n;$ usually denotes a natural number, or at least an integer one, this can be misleading very easily, in my opinion. – DonAntonio Mar 27 '16 at 06:14
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    There are always infinitely many series that sums to any given expression. For example $a_1 = s$ with $a_2=a_3=\ldots=0$ has $\sum a_n = s$. Voting to close as the question has too many answers. – Winther Mar 27 '16 at 06:31
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    Yes it is. For any $k$ let the $k$'th term be $\sqrt{0.5}\sqrt{\frac{1}{1-n}}$ and the rest zero. "The simplest one" is a highly subjective term. The reason is that too broad questions are off-topic on this site. – Winther Mar 27 '16 at 06:52

3 Answers3

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$\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} $

marty cohen
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  • If inside the square root, the result is scaled by $\sqrt{a}$, if outside the square root, scaled by $a$. – marty cohen Mar 27 '16 at 07:03
  • What do you call such a series? it is not geometric it is a sum of square roots, right? so, has it got a particular name? –  Mar 27 '16 at 08:23
  • Nah. It's just a series. I kept on going to try to obfuscate the source as much as possible. It might make an interesting problem to prove thar the final series is equal to $\sqrt(m)$. To make it harder, set $m=2$ to hide the dependence on $m$. – marty cohen Mar 27 '16 at 16:39
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$a_{n+1}=S_{n+1}-S_{n}$ and you have the formula for $S_n$......

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Using Binomial series for $|n|<1,$

$$\sqrt{\dfrac1{1-n}}=(1-n)^{-\frac12}=?$$

The $r+1(\ge0)$th term $$\dfrac{n^r1\cdot3\cdots2r-1}{2^r}=\dfrac{n^r}{4^r}\dfrac{(2r)!}{r!}$$

What happens if $|n|>1?$

lab bhattacharjee
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