$\sum_{n=1}^\infty\frac{n}{(2n-1)16^n}\binom{2n}{n}^2\left(\sum_{k=n}^\infty\frac{2^k}{k\binom{2k}{k}}\right)=1-\sqrt2+\log(1+\sqrt2).$

Prove:$$\sum_{n=1}^\infty\frac{n}{(2n-1)16^n}\binom{2n}{n}^2\left(\sum_{k=n}^\infty\frac{2^k}{k\binom{2k}{k}}\right)=1-\sqrt2+\log(1+\sqrt2).$$

I'm sorry that I don't even know how to start. I haven't met this kind of series before. I've learnt some simplier methods for calculating the value of some simplier series through complex analysis, for example using the Taylor expansion for some holomorphic functions. But it seems that these methods won't work in this problem. So I want to learn more methods or tricks to handle this kind of problems. Would you please recommend some books about this topic?

Introduction

Let me begin by saying that the problem given is problem $4.5$ titled "A Breathtaking Infinite Series Involving the Binomial Coefficient and Expressing a Beautiful Closed-Form" in the book titled "(Almost impossible integrals and sums)" by Cornel Ioan Valean. This astonishing book also contains the proof, in section 6.5. However, the proof is convoluted for a basic level of proficiency in complex analysis. Furthermore, it uses previous results and some standard integrals that are not expounded on in the proof but for which you will need additional sources (which are also present in the book). As I have gauged OP's level from the comments with the help of an example (which will turn out to be of use for me anyway!), I will tailor the following to be a proof of the result, but more than that I will provide background where necessary, and also make remarks regarding how/why the problem was approached in the way it was, and what kind of standard results go into such integral computations.

The standard ideas

See, when solving an integral or summation, one must remember certain theorems and patterns that are very standard tricks in this field. There are many answers on MSE that use these tricks to great effect.

The most basic tricks, those used with little to no justification but often easily justifiable are :

• Simple tricks such as the integration-by-parts, and the use of Taylor series for expansion of functions. IBP leads to recursions which are useful in computing indexed integrals, and Taylor series are important to allow switch of sum and integral, as explained later.

• Interchanging limits and sums : one can exchange limits with sums and integrals under a general condition called the dominated convergence theorem. This result is used often to calculate a finite case and then take a limit to the infinite case.

• Switching multiple summations and integrals simultaneously : using a tool called the Fubini-Tonelli theorem in full generality (so that one can talk about summations as "integrals over the counting measure") one can justify the switching of multiple integrals and summations under either a condition of non-negativity of the summands/integrands, or of their absolute summability/integrability. Here you may read more about these two. Note that in integral computations, one can usually verify absolute S/I with some simple asymptotic analysis (verifying boundedness or sufficient decay at the integral end points if the integrand is continuous, for example) so such steps are not usually expounded on. We will do this for the purpose of being self-contained.

• Interchanging the derivative and the integral : The famous theorem of Lebesgue emphasised here states that one can exchange the partial derivative and integral operations under certain very simple-to-verify conditions. One advantage of this is that often introducing a parameter into an equation and understanding how the integral behaves with respect to that parameter will give great insight into the integral itself. An example of that is illustrated here.

• Complex analysis formulas like the Poisson summation formula, Abel summation formula , the Plancherel identity and so on. These allow us to transition from sums to integrals, by converting a discrete sum into the integral of a continuous quantity formed by playing with the discrete terms.

So, how do we start?

We start, by looking at this inner summation $\sum_{k=n}^\infty \frac{2^k}{k\binom{2k}k}$. It's a sum that we feel can be simplified by using some known results on Taylor series , perhaps it's equal to an integral or something, but it is felt that we should look at it first.

Indeed, what we are going to do first, is rewrite our expression in such a manner , that a recursion reveals itself.

Step 1

Let us recall Wallis' formula , which is going to come in very handy here.

Define for a positive integer $n$, $n!! = n(n-2)...(n \%2)$ where $n\%2 =2$ if $n$ is even and $1$ if $n$ is odd. Then, we have : $$ \int_{0}^{\frac \pi 2} \sin^{n}(t)dt= \begin{cases} \frac{\pi}{2} \frac{(2p-1)!!}{(2p)!!} & n = 2p \\ \frac{(2p)!!}{(2p+1)!!} & n = 2p+1 \end{cases} $$

A proof is available on Wikipedia, but let me give you an exercise to put you on your way.

• Compute $\int_{0}^{\frac \pi 2} \sin(t)dt$ and $\int_{0}^{\frac \pi 2} \sin^2(t)dt$, the latter using the transformation $\cos 2A = 1-2\sin^2 A$.

• Let $n$ be arbitrary, write $W_n = \int_{0}^{\frac {\pi}2} \sin^n(t)dt$. We now rewrite : $$ \int_{0}^{\frac {\pi}2} \sin^n(t)dt = \int_{0}^{\frac {\pi}2} \sin^{n-2}(t)(1-\cos^2(t))dt \\ = \int_{0}^{\frac \pi 2} \sin^{n-2}(t)dt - \int_{0}^{\frac \pi 2} [\cos(t)] [\sin^{n-2}(t)\cos(t)] dt $$

• Integrate the second term by parts using the break-up given, and get back $W_n$, to conclude that $W_n = \frac{n-1}{n}W_{n-2}$.

• Now use recursion to go from $W_n$ to $W_{n-2}$ all the way up to wherever you can go to, and conclude the value of the integral.

What's so handy about Wallis' formula? Well, it's the most elementary example of the find a recursion phenomena. When you see an integral that depends on a parameter, see what happens when the parameter changes and try to see if familiar recursion patterns are available for use. Of course, familiarity comes with experience, therefore this recursion you see here, whenever you see it you have to think of a Wallis integral.

But what of the Wallis integral? Well, we are going to perform a rewrite of our expression, which is going to attempt to involve the familiar $!!$ expression.

Always remember this, in the field of integrals, or rather mathematics, presentation is a very nice facet of problem formulation. Making juice is one thing, but marketing it in a beautifully shaped bottle, with a nice little green label and lovely writing font will make it look tastier.

Something like that has happened here. We need to open the bottle and taste the juice. In the case of mathematics, that's unwrapping the expression and revealing a hidden pattern for recursion.

To bring the $!!$s in , we are going to make two very key observations.

$\frac{(2n-1)!!}{(2n)!!} = \frac{1}{2^{2n}} \binom{2n}{n}$ and $\frac{(2n)!!}{(2n+1)!!} = \frac{2^{2n}}{(2n+1)\binom{2n}n}$.

But how to prove these? Well, let's see the first one. We have : \begin{align} \frac{(2n-1)!!}{(2n)!!}& = \frac{1 \cdot 3 \cdot ... \cdot (2n-1)}{2 \cdot 4 \cdot ... \cdot 2n} \\ &= \frac{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2 \cdot 4 \cdot ...\cdot 2n}{2^2 \cdot 4^2 \cdot ... \cdot (2n)^2 \cdot } \\ &= \frac{(2n)!}{2^{2n} (1 \cdot 2 \cdot ... \cdot n)^2} \\ &= \frac{1}{2^{2n}}\binom{2n}n \end{align}

Bang! The other one follows similarly. Now, we are ready to give our expression a makeover. First, we remove a term which we intend to make a Wallis integral out of. Perhaps you have already seen it yourself. \begin{align} &\frac{n}{(2n-1)16^n} \binom{2n}n^2 \left(\sum_{k=n}^\infty \frac{2^k}{k \binom{2k}k}\right) \\ &= \frac{1}{2^{2n}} \binom{2n}n \left[\frac{n}{(2n-1)2^{2n}} \binom{2n}n \left(\sum_{k=n}^\infty \frac{2^{k}}{k \binom{2k}k}\right)\right] \end{align}

So a Wallis term has come out. Now, you may argue, oh, another Wallis term can come out, can't it? Well, that's a problem because that Wallis term is required for beautifying the other part and get the $!!$s in there. Which comes from trying to force it from the Wallis' formula. $$ \frac{2^{k}}{k \binom{2k}k} = \frac{2^k}{\frac{(2k)!}{k!(k-1)!}} = \frac{2^k k (k-1)!^2}{(2k)(2k-1)(2k-2)!} = \frac{2^{k-1}}{(2k-1)\binom{2k-2}{k-1}} = \frac{1}{2^{k-1}} \frac{(2k-2)!!}{(2k-1)!!} $$ and therefore : $$ \sum_{k=n}^\infty \frac{2^{k-1}}{k \binom{2k}k} = \sum_{k=n}^\infty \frac{1}{2^{k-1}} \frac{(2k-2)!!}{(2k-1)!!} $$ Similarly for the stuff outside : $$ \frac{n}{(2n-1)2^{2n}} \binom{2n}n = \frac{n}{(2n-1)}\frac{(2n-1)!!}{(2n)!!} = \frac{n \cdot (2n-1)\cdot (2n-3)\cdot ... \cdot 1}{(2n) \cdot (2n-1) \cdot (2n-2) \cdot (2n-4)\cdot ... \cdot 2} \\ = \frac{(2n-3)!!}{2(2n-2)!!} $$ Finally, taking the $2$ in this denominator inside the summation to make the $2^{k-1}$ into a $2^k$ gives : $$ \frac{n}{(2n-1)2^{2n}} \binom{2n}n \left(\sum_{k=n}^\infty \frac{2^{k}}{k \binom{2k}k}\right) = \frac{(2n-3)!!}{(2n-2)!!}\sum_{k=n}^\infty \frac{1}{2^{k}} \frac{(2k-2)!!}{(2k-1)!!} $$

There! That's better. Now, we have got this whole lot of double factorials to deal with. And how are we supposed to do that?

Step 2

Step $2$ comes from thinking : ah, ok , so we've got something Wallis-like, by the looks of it. I should THINK of some integral involving sine and cosine powers, maybe? And then perhaps Wallis will help me sort things out?

Well, you've struck gold if you thought that way!

Let's think about the term $$ \frac{(2n-3)!!}{(2n-2)!!}\sum_{k=n}^\infty \frac{1}{2^{k}} \frac{(2k-2)!!}{(2k-1)!!} $$ move $n$ forward by $1$ to get : $$ \frac{(2n-1)!!}{(2n)!!}\sum_{k=n+1}^\infty \frac{1}{2^{k}} \frac{(2k-2)!!}{(2k-1)!!} $$ Now, we are going to find the relation between these terms, in terms of a recursion, and try to match that with a Wallis recursion, if we can see it. \begin{align} \frac{(2n-1)!!}{(2n)!!}\sum_{k=n+1}^\infty \frac{1}{2^{k}} \frac{(2k-2)!!}{(2k-1)!!}& = \frac{2n-1}{2n} \left[\frac{(2n-3)!!}{(2n-2)!!}\sum_{k=n+1}^\infty \frac{1}{2^{k}} \frac{(2k-2)!!}{(2k-1)!!}\right] \\ &= \frac{2n-1}{2n}\left[\frac{(2n-3)!!}{(2n-2)!!}\sum_{k=n}^\infty \frac{1}{2^{k}} \frac{(2k-2)!!}{(2k-1)!!}\right] - \frac{(2n-1)(2n-3)!!(2n-2)!!}{(2n)(2n-2)!!2^n(2n-1)!!} \\ &= \frac{2n-1}{2n}\left[\frac{(2n-3)!!}{(2n-2)!!}\sum_{k=n}^\infty \frac{1}{2^{k}} \frac{(2k-2)!!}{(2k-1)!!}\right] - \frac{1}{n2^{n+1}} \end{align}

So we are very much reminded of the Wallis recurrence, are we not? If I call $$S_{2n} = \frac{(2n-1)!!}{(2n)!!}\sum_{k=n+1}^\infty \frac{1}{2^{k}} \frac{(2k-2)!!}{(2k-1)!!}$$ then the above recurrence is $S_{2n} = \frac{2n-1}{2n}S_{2n-2} - \frac 1{n2^{n+1}}$, which without that little boundary term, is the Wallis integral!

But that $\frac 1{n2^{n+1}}$ kind of sticks out like a sore thumb. What to do with it? Well, recall integration by parts. Remember that "boundary -term", which vanishes when we compute the Wallis integral? Well, we have to now incorporate that term as an integration-by-parts boundary term.

Which means that the boundaries change. But what should they be? Well, recall that the boundary term in the Wallis integral is $\frac{\sin^{n-1}(t)\cos(t)}{n-1}$ evaluated at the boundary points. Can we think of two angles, such that the difference of this quantities evaluated at the two angles is $\frac 1{n2^{n+1}}$? Well, the answer comes from noting that if $t= \frac \pi 4$ then that term is $\frac{1}{n2^{n-1}}$, which immediately tells us : OH, Wallis integral, but with limits $0$ and $\frac\pi 4$!

Exercise : Let $I_{2n} = \int_0^{\frac \pi 4} \sin^{2n}(t)dt$. Prove, using the technique used for the Wallis formula, that : $$ I_{2n} =\frac{2n-1}{2n}I_{2n-2} - \frac{1}{n2^{n+1}} $$

But wait : isn't this the exact same recurrence as $S_{2n}$? So surely we must have $I_{2k} = S_{2k}$ for all $k$?

Well, we'd be done, except we actually need to ensure $S_2 = I_2$. Or at least, see the difference between the two. Then we can actually find and express $S_{2n}$ as a Wallis integral.

Well, $I_2$ is easily calculated using the double angle formula, it is $\frac{\pi}{8} - \frac 14$. So, we'd expect $S_2$ to equal this. What would happen if it were equal? $$ S_2 = \frac 12 \sum_{k=2}^\infty \frac{(2k-2)!!}{2^k(2k-1)!!} = \frac{\pi}{8} - \frac 14 \\ \implies \sum_{k=1}^\infty \frac{(2k-2)!!}{2^k(2k-1)!!} = \frac \pi 4 $$

(Note : $0!! = 1$ is used here, it is accepted as convention) And thus, another step comes from the question : is the above true?

Step 3

Well, it is true! What's the idea, you may ask. Well, the idea comes from the fact that the double factorials are representable as Wallis integrals. Therefore, combining with the $2^k$ term, one gets a geometric series, and using an exchange of integral and sum one can easily obtain something tangible.

Let's see how it works now. $$ \sum_{k=1}^\infty \frac{(2k-2)!!}{2^k(2k-1)!!} = \sum_{k=1}^\infty \frac {1}{2^k} \frac{(2k-2)!!}{(2k-1)!!} = \sum_{k=1}^\infty \frac 1{2^k} \int_{0}^{\frac \pi 2} \sin^{2k-1}(t)dt $$

We exchange the sum and the integral (proving this from Fubini is not very difficult) and we adjust the $\sin$ term a little to see where a geometric series comes from : $$ \sum_{k=1}^\infty \frac 1{2^k} \int_{0}^{\frac \pi 2} \sin^{2k-1}(t)dt =\int_0^{\frac \pi 2} \frac 1{\sin t}\sum_{k=1}^\infty \frac{\left(\sin(t)^{2}\right)^k}{{2}^{k}}dt = \int_{0}^{\frac \pi 2} \frac 1{\sin t} \frac{\frac{\sin^2 t}{2}}{1 - \frac{\sin^2 t}{2}} dt $$

and with a few cancellations : $$ \int_{0}^{\frac \pi 2} \frac 1{\sin t} \frac{\frac{\sin^2 t}{2}}{1 - \frac{\sin^2 t}{2}} dt = \int_{0}^{\frac {\pi} 2} \frac{\sin t}{2 - \sin^2 t} dt $$

That latter integral is quite easy, in comparison to what we've done so far. Can you prove that it is equal to $\frac \pi 4$? I leave this as an exercise for you, with two hints provided : $$ \frac{\sin t}{2 - \sin^2 t} = \frac{\sin t}{1+\cos^2 t} $$ , and the derivative of the arctangent function. Now go for it.

This proves, remarkably, that $S_{2} = I_2$, and therefore, we have a Wallis' integral type formulation for the complicated expression that we had to deal with.

Step 4

Let's finish now, by inserting everything in, and seeing what we have to do.

\begin{align} &\sum_{n=1}^\infty \frac{n}{(2n-1)16^n} \binom{2n}n^2 \left(\sum_{k=n}^\infty \frac{2^k}{k \binom{2k}k}\right) \\ &= \sum_{n=1}^\infty\frac{1}{2^{2n}} \binom{2n}n \left[\frac{n}{(2n-1)2^{2n}} \binom{2n}n \left(\sum_{k=n}^\infty \frac{2^{k}}{k \binom{2k}k}\right)\right] \\ &= \sum_{n=1}^\infty \left[\frac 2 \pi\int_0^{\frac \pi 2} \sin^{2n}(t)dt\right] \left[\int_0^{\frac \pi 4} \sin^{2n-2}(x)dx\right] \end{align}

Slide the sum in (in this case, all the integrands are non-negative so no issue!), adjust to get a geometric series, sum it. \begin{align} &\sum_{n=1}^\infty \left[\frac 2 \pi\int_0^{\frac \pi 2} \sin^{2n}(t)dt\right] \left[\int_0^{\frac \pi 4} \sin^{2n-2}(x)dx\right] \\ &= \frac 2 \pi\int_{0}^{\frac \pi 2} \int_{0}^{\frac \pi 4} \frac 1{\sin^2 x} \sum_{n=1}^\infty \sin^{2n}(t)\sin^{2n}(x)dxdt \\ &= \frac 2 \pi\int_{0}^{\frac \pi 2} \int_{0}^{\frac \pi 4} \frac 1{\sin^2 x} \frac{\sin^2 t \sin^2 x}{1 - \sin^2 t \sin^2 x}dxdt \\ &= \frac 2 \pi\int_{0}^{\frac \pi 2} \int_{0}^{\frac \pi 4} \frac{\sin^2 t}{1 - \sin^2 t \sin^2 x}dxdt \end{align}

We are now left with a double integral to evaluate. So we do the usual thing ; do the inner integral first. So we treat $\sin^2(t)$ like a constant, say $\sin^2 t = b$. So the inner integral is : $$ \int_{0}^{\frac \pi 4} \frac{b}{1 - b\sin^2 x}dx $$

A little bit of famililarity will tell you that there is a lovely trick here, a school like trick. $$ \int_{0}^{\frac \pi 4} \frac{b}{1 - b\sin^2 x}dx = b\int_0^{\frac \pi 4} \frac{\sec^2 x}{\sec^2 x - b \tan^2 x}dx = b \int_0^{\frac \pi 4} \frac{\sec^2 x}{1 + (1-b)\tan^2 x} dx $$

Set $u = \tan x$ so $du = \sec^2 x dx$. We get : $$ b \int_0^{\frac \pi 4} \frac{\sec^2 x}{1 + (1-b)\tan^2 x} dx = b\int_0^1 \frac{du}{1+(1-b)u^2} $$ Putting $b = \sin^2 t$ back gives : $$ \frac 2 \pi\int_{0}^{\frac \pi 2} \int_{0}^{\frac \pi 4} \frac{\sin^2 t}{1 - \sin^2 t \sin^2 x}dxdt = \frac 2 \pi\int_{0}^{\frac \pi 2} \int_{0}^{1} \frac{\sin^2 t}{1 +u^2 \cos^2 t}dudt $$

We exchange the integrals : $$ \frac 2 \pi\int_{0}^{\frac \pi 2} \int_{0}^{1} \frac{\sin^2 t}{1 +u^2 \cos^2 t}dudt = \frac 2 \pi\int_{0}^{1} \int_{0}^{\frac \pi 2} \frac{\sin^2 t}{1 +u^2 \cos^2 t}dtdu $$

For the inner integral, we invoke an identity I wish you verify. Let $l,m$ be real numbers with $m,l \neq 0$, $ml \neq -1$, then : $$ \frac{1-l}{1+ml} = \frac{1}{1+ml} - \frac{l}{1+ml} = \frac{1}{1+ml} - \frac 1m\frac{ml}{1+ml} = \frac {1}{1+ml} - \frac 1m\left(1 - \frac{1}{1+ml}\right) \\ = \frac{1}{1+ml}\left( 1 + \frac 1m\right) - \frac 1m $$

using this identity appropriately gives : $$ \frac{\sin^2 t}{1 +u^2 \cos^2 t} = \frac{1-\cos^2 t}{1 +u^2 \cos^2 t} = \frac 1{1 +u^2 \cos^2 t} \left(1 + \frac 1{u^2}\right) - \frac 1{u^2} $$

inserting the integral sign and gives : $$ \int_{0}^{\frac \pi 2} \frac{\sin^2 t dt}{1 +u^2 \cos^2 t} = \left(1 + \frac 1{u^2}\right)\int_0^{\frac \pi 2} \frac {dt}{1 +u^2 \cos^2 t} - \frac{\pi}{2}\frac 1{u^2} $$

The integral $\int_{0}^{\frac \pi 2} \frac {dt}{1 +u^2 \cos^2 t}$ is easily evaluated using a $\tan$ transformation : multiply top and bottom by $\sec^2 t$, express the denominator purely using $\tan t$ and substitute $v = \tan t$. As an exercise, I leave you to see using this heuristic that : $$ \left(1 + \frac 1{u^2}\right)\int_0^{\frac \pi 2} \frac {dt}{1 +u^2 \cos^2 t} - \frac{\pi}{2}\frac 1{u^2} = \frac{\pi}{2}\left(1 + \frac 1{u^2}\right) \frac{1}{\sqrt{1+u^2}} - \frac{\pi}{2}\frac 1{u^2} $$

so we get following the $\frac \pi 2$ cancelling : $$ \frac 2 \pi\int_{0}^{1} \int_{0}^{\frac \pi 2} \frac{\sin^2 t}{1 +u^2 \cos^2 t}dtdu = \int_0^1 \left[\left(1 + \frac 1{u^2}\right) \frac{1}{\sqrt{1+u^2}} - \frac 1{u^2} \right] du $$

We break this into three parts, each of which are of varying difficulty to solve but still easy enough for someone even reasonably well versed in integrals : $$ \int_0^1 \left[\left(1 + \frac 1{u^2}\right) \frac{1}{\sqrt{1+u^2}} - \frac 1{u^2} \right] du = \int_0^1 \frac{1}{\sqrt{1+u^2}}du - \int_0^1 \frac{1}{u^2}\left(\frac{1}{\sqrt{1+u^2}} - 1\right)du $$

The first is a proper integral matching with the $\mbox{arcsinh}$ integral, which gives you $\mbox{arcsinh}(1)$. But , $\mbox{arcsinh}(x) =\log(x + \sqrt{1+x^2})$ (this is a quite easily verified formula) so that term is $\log(1+\sqrt 2)$.

The second term is an improper integral, so we look at $\int_{\epsilon}^1$ of the same quantity and take the limit. That is, we have : $$ \int_0^1 \frac{1}{u^2}\left(\frac{1}{\sqrt{1+u^2}} - 1\right)du = \lim_{\epsilon \to 0} \int_{\epsilon}^1 \frac{1}{u^2}\left(\frac{1}{\sqrt{1+u^2}} - 1\right)du $$

Now, I leave you to break the integral into the obvious parts and prove that : $$ \int \frac{1}{u^2}\left(\frac{1}{\sqrt{1+u^2}} - 1\right)\,du = \frac{-\sqrt{u^2+1}+1}{u}+C $$

So : $$ \int_{\epsilon}^1 \frac{1}{u^2}\left(\frac{1}{\sqrt{1+u^2}} - 1\right)du = -\sqrt 2 +1 - \frac{\sqrt{\epsilon^2+1}-1}{\epsilon} $$

The limit of the last quantity as $\epsilon \to 0$ is zero. So you get a $1 - \sqrt 2$ from here.

Finally, the answer $1 - \sqrt 2 + \log(1+\sqrt {2})$ comes out, finishing the ordeal.

This is not an answer but it is too long for a comment.

Looking at the lhs, I was expecting a lot of possible things expect what is the rhs. $$\sum_{k=n}^\infty\frac{2^k}{k \binom{2 k}{k}}=\frac{2^n }{n \binom{2 n}{n}}\,\, _2F_1\left(1,n;n+\frac{1}{2};\frac{1}{2}\right)$$ where appears the gaussian hypergeometric function.

The summand is then $$a_n=\frac{ \binom{2 n}{n} }{(2n-1)\, 8^n}\,\, _2F_1\left(1,n;n+\frac{1}{2};\frac{1}{2}\right)$$ $$a_n=(-1)^n \frac{ \binom{-\frac{1}{2}}{n}}{2^n\,(2 n-1)}\,\, _2F_1\left(1,n;n+\frac{1}{2};\frac{1}{2}\right)$$

and, for any $n$ (this just comes from the hypergeometric function) $$a_n=\alpha _n +\beta_n \pi$$ were the $\alpha _n$ and $\beta_n$ are rational numbers. For example, $$\sum_{n=1}^{10} a_n=\frac{47 (134765953875 \pi -208227582208)}{21646635171840}$$ which is $0.467147$ while the value of the rhs is $0.4671600$.

I do not see how to exploit the fact that $$\, _2F_1\left(1,n;n+\frac{1}{2};\frac{1}{2}\right)=\sum_{p=1}^{\infty}\frac{ \left(n-\frac{1}{2}\right)! \,(n+p-1)!}{2^p\,(n-1)! \left(n+p-\frac{1}{2}\right)!}$$

At this point, I am completely stuck.