Question

Show that
\[
\sum _{n=1}^{\infty } \frac{(2 n-1)!}{2^{4 n}
(n!)^2}=\ln \left(-4
\left(\sqrt{3}-2\right)\right)
\]

Answer 1

Knowing the above property, one can give a nice solution to
http://openstudy.com/study#/updates/54a2cb99e4b054f0c3b448b0

@ganeshie8 @cambrige @SithsAndGiggles

Answer 2

Is there any chance we could write the LHS as a power series of a known function with \(x=\dfrac{1}{2}\)?

Answer 3

Or \(x=\dfrac{1}{16}\)?

Answer 4

Just at first glance I recognized that we can rewrite some of this with even and odd factorial functions but after making some soup I don't know what else I can really do.

Answer 5

Soup strikes again...

Answer 6

My only other strategy is turn lnx into a power series and then play around with the terms and try to make them look alike but... It doesn't look very promising lol.

Answer 7

It somewhat resembles the Bessel function of the first kind: http://en.wikipedia.org/wiki/Bessel_function#Bessel_functions_of_the_first_kind_:_J.CE.B1

if it weren't for the \((2n-1)!\) in the numerator.

Answer 8

Or rather the modified version: http://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions_:_I.CE.B1.2C_K.CE.B1

Answer 9

I've been studying tensors all morning and I'm burnt out on this kind of hard stuff haha. I'm gonna try to help some kids with slope intercept lol

Answer 10

Hyperbolic trig perhaps?

Answer 11

We have a winner: http://en.wikipedia.org/wiki/Inverse_hyperbolic_function#Series_expansions

Answer 12

Wow, that makes the natural logarithm part make sense quite a bit, now that I see that, I should have thought that earlier, good call.

Answer 13

So if
\[\newcommand{\arcsech}{\text{arcsech}}
\arcsech x=\ln\frac{2}{x}-\sum_{n=1}^\infty \frac{(2n-1)!}{2^{2n}(n!)^2}x^{2n}\]
then
\[\arcsech \frac{1}{2}=\ln4-\sum_{n=1}^\infty \frac{(2n-1)!}{2^{4n}(n!)^2}~~\iff~~\sum_{n=1}^\infty \frac{(2n-1)!}{2^{2n}(n!)^2}x^{2n}=\ln4-\ln(2+\sqrt3)\]

Answer 14

... and some minor algebraic manipulation seals the deal! Great question!

Answer 15

\[\arcsech \frac{1}{2}=\ln4-\sum_{n=1}^\infty \frac{(2n-1)!}{2^{4n}(n!)^2}~~\iff~~\sum_{n=1}^\infty \frac{(2n-1)!}{2^{\color{red}{4n}}(n!)^2}=\cdots\]