# Compute the series : $\sum_{n=1}^{\infty} \frac{4^n n!}{(2n)!}$

How would you compute the following series? I'm interested in some easy approaches that would allow me to work it out. $$\sum_{n=1}^{\infty} \frac{4^n n!}{(2n)!}$$

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Try the Laplace transform approach. – Mhenni Benghorbal Feb 21 '13 at 16:37
@MhenniBenghorbal: Does Laplace work here? Interesting! Thank you for your precious feedback! – user 1618033 Feb 21 '13 at 16:44
I have not worked out. Just try to follow the technique and see what you get. I'll try to do this later. – Mhenni Benghorbal Feb 21 '13 at 16:47
@MhenniBenghorbal: I saw the post. It's really precious that way. – user 1618033 Feb 21 '13 at 18:18

It suffices to calculate the sum with the summation index running from 0, not from 1. Then

\begin{align*} \sum_{n=0}^{\infty} \frac{4^n n!}{(2n)!} &= \sum_{n=0}^{\infty} \frac{4^n}{(2n)!}\int_{0}^{\infty}t^n e^{-t}\;dt\\ &= \int_{0}^{\infty}\sum_{n=0}^{\infty} \frac{4^n t^n}{(2n)!} e^{-t}\;dt\\ &= \int_{0}^{\infty} \cosh(2\sqrt{t}) e^{-t}\;dt\\ &= \int_{0}^{\infty} 2u\cosh(2u) e^{-u^2}\;du\qquad(t=u^2)\\ &= \left[-\cosh(2u)e^{-u^2}\right]_{0}^{\infty} + 2\int_{0}^{\infty} \sinh(2u) e^{-u^2}\;du\\ &= 1 + 2\int_{0}^{\infty} \sinh(2u) e^{-u^2}\;du \\ &= 1 + \int_{0}^{\infty} \left(e^{2u} - e^{-2u}\right) e^{-u^2}\;du \\ &= 1 + e \int_{0}^{\infty} \left(e^{-(u-1)^2} - e^{-(u+1)^2}\right)\;du. \end{align*}

Now it is clear how to express the last integrals in terms of the error function, yielding

$$\sum_{n=1}^{\infty} \frac{4^n n!}{(2n)!} = e\sqrt{\pi} \mathrm{erf}(1).$$

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+1 nice   – draks ... Aug 7 '12 at 6:56
Good answer! Thanks! – user 1618033 Aug 7 '12 at 7:03

Here is a generalization. Consider the generating function

$$f(x) = \sum_{n \ge 1} \frac{x^{2n} n!}{(2n)!}.$$

It's an easy check that $2$ is in the radius of convergence, so that we need to compute $f(2)$ to solve your problem. However, it's easy to see that

$$f''(x) = 1 + \sum_{n \ge 1} \frac{x^{2n} (n + 1)!}{(2n)!}$$

Moreover,

$$xf'(x) = 2 \sum_{n \ge 1} \frac{x^{2n} n!}{(2n)!} \cdot n$$

We get the second-order ODE

$$\frac{xf'(x)}{2} + f(x) = f''(x) - 1, f(0) = 0, f'(0) = 0.$$

And the solution

$$f(x) = \frac{1}{2} \cdot e^{x^2/4} \cdot x \sqrt{\pi} \mathrm{erf}(x/2).$$

Just plug in $x = 2$ to get your answer, $e\sqrt{\pi} \mathrm{erf}(1)$.

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mersi fain pentru solutia oferita. (Thank you!) – user 1618033 Aug 7 '12 at 13:25

This is not an appropriate answer. We have $n!\approx C n^{n+1/2}e^{-n}$, so the terms become approximate $$4^{n}\frac{C n^{n}n^{1/2}e^{-n}}{C 2^{2n}2^{1/2}n^{2n}n^{1/2}e^{-2n}}$$

After cancelling out the terms we should have $$\frac{1}{2^{1/2}e^{-n}n^{n}}=2^{-1/2}(\frac{e}{n})^{n}$$

I do not know if there is any way to sum $$C\sum^{\infty}_{N}(\frac{e}{n})^{n}$$ where $N$ is some very large number depends on Stirling's formula. Now suppose we have $kn,0\le k\le 1$ instead of $e$. Then we should have $$\sum^{\infty}_{N}k^{n}=\frac{k^{N}}{1-k}$$

So obviously for fixed $N$ we have the top side of the derivative to be $$Nk^{N-1}(1-k)+k^{N}=Nk^{N-1}-(N-1)k^{N}$$ which equals 0 if and only if $$k=\frac{N}{N-1}$$ which is impossible by our assumption on $k$. Thus this sum increases monotone with $k$ and has a minimum at $k=0$. So assume $k=\frac{e}{N}$ we should be able to bound the original sum from above. For the exact value one probably need to run a computer program.

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