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show that: $$\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\left(\frac{1}{2}\right)^{1/2}?$$

this sum is from other problem,if I solve this,then the other problem is solve it

mookid
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    Show that $$\frac 1{\sqrt{1-x}}=\sum_{n=0}^\infty \frac{(2n)!}{n!^2}\left(\frac x4\right)^n$$ For other central-binomial series see pi314.net – Raymond Manzoni Oct 27 '14 at 14:13
  • I do not find this result. See my answer (there could very well be an overlook in my answer). – mookid Oct 27 '14 at 14:51
  • @mookid: comparing expansions with pari/gp : $$\sqrt{1-x}= 1 - x/2 - x^2/8 - x^3/16 - 5x^4/128-\cdots$$
    $$1/\sqrt{1-x}= 1 + x/2 + 3x^2/8 + 5x^3/16 + 35x^4/128+\cdots$$

    vector(6,n,binomial(2*n,n)/4^n) = [1/2, 3/8, 5/16, 35/128, 63/256, 231/1024]
    Anyway the derivative of the first one should be proportional to the second one...

     – Raymond Manzoni Oct 27 '14 at 15:01
  • @RaymondManzoni I don't see the bug. I don't have pari either. If you know where it is, thanks for pointing it out! – mookid Oct 27 '14 at 15:13

3 Answers3

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Hint: what is $\frac{f^{(n)}(0)}{n!}$ when $f(x) = \sqrt{1-x}$?

Compute the first terms: \begin{align} f^{(0)}(x) &= \sqrt{1-x}\\ f^{(1)}(x) &= -\frac 12 (1-x)^{-1/2} \\ f^{(2)}(x) &= -\frac 14 (1-x)^{-3/2} \\ \end{align}

The exponent of $1-x$ in $f^{(n)}(x)$ is obviously $1/2 - n$. So when you derivate, the front factor gets multiplied by $1/2 - n$ to get $f^{(n+1)}(x)$, and by $-1 $ because of the chain rule.

Hence the result is \begin{align}f^{(n+1)}(0) &= 1\times \left(\frac 12 - 0\right) \times \left(\frac 12 - 1\right) \times \dots \times \left(\frac 12 - n\right) \times (-1)^{n+1} \\ &= \frac{1\times 3 \times \dots \times (2n-1)}{2^{n+1}} (-1)^n \times (-1)^{n+1} = -\frac{1\times 3 \times \dots \times (2n-1)}{2^{n+1}}\\ &= -\frac{(2n)!}{2^{n+1} 2\times 4 \times \dots \times 2n} = -\frac{(2n)!}{2^{2n+1} n!} \end{align}

You can check that this is true for the first values: \begin{align} f^{(1)}(x) &= -\frac 12 = 1\times \frac 12 &\times (-1)^1 \\ f^{(2)}(x) &= -\frac 14 = 1\times \frac 12 \times \left(-\frac 12\right) &\times (-1)^2 \end{align}

Then, you must show that the series is convergent for $x=\frac 12$ and you will get: \begin{align} \sqrt{1-x} &= 1 + \sum_{n=0}^\infty \frac{f^{(n+1)}(0) }{(n+1)!}x^{n+1} \\ &= 1 - \sum_{n=0}^\infty \frac{(2n)!}{2^{2n+1} (n!)^2 (n+1)} x^{n+1} \end{align}

Unfortunately, when evaluated in $\frac 12$, this does not make appear the desired sum, because of the factor $n+1$.

We can get rid of it just using tern by term derivation: \begin{align} -\frac 12 (1-x)^{-1/2} &=- \sum_{n=0}^\infty \frac{(2n)!}{2^{2n+1} (n!)^2} x^n\\ (1-x)^{-1/2} &= \sum_{n=0}^\infty \frac{(2n)!}{2^{2n} (n!)^2} x^n \end{align}

And now we get, with $x=1/2$: $$ \left(1-\frac 12\right)^{-1/2} = \sum_{n=0}^\infty \frac{(2n)!}{2^{3n} (n!)^2} $$

mookid
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  • I believe that something is amiss; the sum is $\sqrt{\frac12}$. – robjohn Oct 27 '14 at 15:16
  • I don't find the error here, even if the result is wrong. – mookid Oct 27 '14 at 15:22
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    I get $$(1-x)^{1/2}=1-\sum_{k=0}^\infty\frac{(2k)!}{2^{2k+1}k!^2}\frac{x^{k+1}}{k+1}$$ You are dividing by $n!$ instead of $(n+1)!$ on the $x^{n+1}$ term, it seems. – robjohn Oct 27 '14 at 15:30
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    Shouldn't your denominator in $$\sum_{n=0}^\infty \frac{f^{(n+1)}(0) }{n!} \left(\frac 12\right)^{n+1}$$ be $(n+1)!;$ so that the derivative should give the correct result ? (robjohn found the same thing it seems! :-)) – Raymond Manzoni Oct 27 '14 at 15:36
  • @RaymondManzoni thanks! I'll see if I can fix this easily. – mookid Oct 27 '14 at 17:04
  • @robjohn thanks! It seems I get the wrong sum... I think I should delete this message. – mookid Oct 27 '14 at 17:07
  • @mookid: It might be difficult to approach the answer using this function, but it is possible that your answer might be salvaged. If you can salvage it, that would be a benefit to the site. If you can't salvage it, then that indicates another path. – robjohn Oct 27 '14 at 18:29
  • @robjohn An easy fix is possible, I added it in the end. Thanks again for tracking the error! – mookid Oct 27 '14 at 18:42
  • @mookid: somewhere in there, a factor of $2$ was dropped. The left side of the last line is $\sqrt2$ when it should be $\sqrt{\frac12}$. You have $$\sqrt{1-x} = 1 - \sum_{n=0}^\infty \frac{(2n)!}{2^{2n+2} (n!)^2 (n+1)} x^{n+1}$$ but the right side does not look like $1-\frac12x+\dots$ – robjohn Oct 27 '14 at 19:13
  • looks much better now :-) – robjohn Oct 27 '14 at 19:18
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Hint: Using the generalized binomial theorem, we get $$ \begin{align} \left(1-x\right)^{-1/2} &=1+\frac12x+\frac{\frac12\cdot\frac32}{2!}x^2+\frac{\frac12\cdot\frac32\cdot\frac52}{3!}x^3+\frac{\frac12\cdot\frac32\cdot\frac52\cdot\frac72}{4!}x^4+\dots\\[6pt] &=\sum_{k=0}^\infty\frac{(2k-1)!!}{2^kk!}x^k\\ &=\sum_{k=0}^\infty\frac{(2k)!}{2^{2k}k!^2}x^k\\ \end{align} $$ Plug in $x=\frac12$ and check the result.

robjohn
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This is a binomial series. Try to rewrite $\bigg(\dfrac12\bigg)^\tfrac12$ as $\bigg(1-\dfrac12\bigg)^\tfrac12$ or as $\Big(1+1\Big)^{^{-\tfrac12}}$. Then expand, simplify the general term, and see what happens. :-$)$

Lucian
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