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How does one sum the given series: $$ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{2n+5} + \frac{ 1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{1}{2n+7} + \cdots \ \text{ad inf}$$

Given, such a series, how does one go about solving it. Getting an Integral Representation seems tough for me.

I thought of going along these lines, but couldn't succeed.

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up vote 10 down vote accepted

Observe that

$$\displaystyle \frac{1}{4^n} {2n \choose n} = \frac{(2n-1)(2n-3)(2n-5)...}{2n(2n-2)(2n-4)...}$$

and recall that

$$\displaystyle \frac{1}{ \sqrt{1 - x^2} } = \sum_{k \ge 0} \frac{1}{4^k} {2k \choose k} x^{2k}.$$

It follows that the desired quantity is

$$\displaystyle \int_0^1 \frac{x^{2n}}{\sqrt{1 - x^2}} dx.$$

But letting $x = \sin \theta$ this is just

$$\displaystyle \int_0^{ \frac{\pi}{2} } \sin^{2n} \theta d \theta = \frac{\pi}{2} \frac{1}{4^n} {2n \choose n}.$$

This is equivalent to Mariano's closed form via the identities $\Gamma(x+1) = x \Gamma(x)$ and $\Gamma \left( \frac{1}{2} \right) = \sqrt{\pi}$. I should mention that the integral at the end of Moron's answer is quite doable; write it in terms of a cosine and use the substitution $\cos \theta = \frac{1 - t^2}{1 + t^2}$ where $t = \tan \frac{\theta}{2}$ to reduce the problem to the integral of a rational function, and then one can use one of several related methods (partial fractions, contour integration).

Remark: The last integral identity above happens to be one of my favorite identities. I describe a representation-theoretic and combinatorial proof of it in this blog post. Another approach implicitly occurs in this blog post.

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@Q.Yuan: Great idea! Hey what made you to look at this : $\frac{1}{4^n} {2n \choose n}$ – anonymous Sep 1 '10 at 6:12
It's the natural expression you get if you try to write everything in terms of factorials. But I have also seen this identity before; additional details will be edited into the answer. – Qiaochu Yuan Sep 1 '10 at 6:20

Consider the function

$$f(a) = \sum_{k=0}^{\infty} \frac{1\cdot3\cdots(2k-1)}{2\cdot4\cdots(2k)}\cdot a^{2n+2k}$$

$$|a| \leq 1$$

What you want is $$\int_{0}^{1} f(a) da$$

The $k^{th}$ coefficient can be written as


(again Wallis's like product as in the other question).

Thus we have

$$f(a) = \frac{2a^{2n}}{\pi} \int_{0}^{\frac{\pi}{2}} \sum_{k=0}^{\infty} {(a\sin x)^{2k}} dx$$

$$ = \frac{2a^{2n}}{\pi} \int_{0}^{\frac{\pi}{2}} {\frac{1}{1-(a\sin x)^2}}dx$$

Now $$\frac{1}{1-(a\sin x)^2} = \frac{1}{\cos^2 x + (\sqrt{1-a^2}\sin x)^2} = \frac{\sec^2 x}{1+(\sqrt{1-a^2}\tan x)^2}$$ which is the derivative of

$$\frac{\arctan(\sqrt{1-a^2}\tan x)}{\sqrt{1-a^2}}$$ whose integral between $0$ and $\frac{\pi}{2}$ is $$\frac{\pi}{2\sqrt{1-a^2}}$$

Thus we get $$f(a) = \frac{a^{2n}}{\sqrt{1-a^2}}$$

(So in fact, we have shown the identity which Qiaochu uses in his answer).

Hence the required sum is $$\int_{0}^{1} \frac{a^{2n}}{\sqrt{1-a^2}}da$$ which can easily be found by the substitution $a = \sin x$ (like in Qiaochu's answer) and which gives us the result to be

$$\int_{0}^{\frac{\pi}{2}} {\sin^{2n}x}dx$$

Now this integral for positive integer $n$ can easily be seen to be what Mariano got. I am unsure of whether that formula holds for an arbitrary real number $n$.

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Mathematica computes your double integral into $$\frac{\sqrt{\pi } \Gamma\left(n+\frac{1}{2}\right)}{2 \Gamma (n+1)}$$, which equals the sum of the series it gives. – Mariano Suárez-Alvarez Sep 1 '10 at 5:34
Dude, it's like you memorized the Wallis product formula... :) – J. M. Sep 1 '10 at 6:32
@J.M: :-) The last problem kind of reminded me, and by looking that up, I came across this one too. – Aryabhata Sep 1 '10 at 13:59

FWIW, Mathematica is able to sum this:

In[1]:= Sum[1/(2 n + 2 k + 1) (2 k - 1)!!/(2 k)!!, {k, 0, \[Infinity]}]

                        3 + 2 n
        Sqrt[Pi] Gamma[-------]
Out[1]= -----------------------
        (1 + 2 n) Gamma[1 + n]
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