The series $$\sum_{n=1}^\infty \frac{(2n)!}{(n!)^2}$$ Is divergent. Is this statement true?
I tried by Root Test.
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Martin Sleziak
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user458361
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4The ratio test is simpler here. – Daniel Fischer Jun 27 '17 at 20:26
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3Does the summand go to zero as $n\to \infty$? – User8128 Jun 27 '17 at 20:26
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$n\mapsto\binom{2n}{n}$ is quite blatantly an increasing and unbounded function. Maybe you mean $$ \sum_{n\geq 1}\frac{(2n)!}{4^n (n!)^2} = +\infty, $$ that is less trivial. – Jack D'Aurizio Jun 27 '17 at 20:46
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The terms $\binom{2n}{n}= \frac{(2n)!}{(n!)^2}$ in this summation are called central binomial coefficients because in each row of Pascal's triangle, the middle entry is the largest.
Rather than approaching zero (as convergence of a series requires), these terms grow roughly like $4^n$. A more precise estimate is given by the Wikipedia article (linked above) and proved to be an upper bound in this previous MathOverflow Question:
$$ \binom{2n}{n} \approx \frac{4^n}{\sqrt{\pi n}} $$
Indeed the related power series $\sum_{n=0}^\infty \binom{2n}{n} x^n$ converges to $\frac{1}{\sqrt{1-4x}}$ for $x\in [-1/4,1/4)$. One might suspect that the original form of the problem that inspired this Question was to determine the radius of convergence of this power series (which is $R=1/4$ by location of the nearest singularity to the origin).
hardmath
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Yes, this series is clearly divergent because each term it does not approach $0$ as $n$ approaches $\infty$. If we split up $$\frac{(2n)!}{(n!)^2}$$ like this: $$\frac{n!\cdot(n+1)(n+2)(n+3)...(2n)}{n!n!}$$ $$\frac{\cdot(n+1)(n+2)(n+3)...(2n)}{n!}$$ $$\frac{n+1}{1}\cdot\frac{n+2}{2}\cdot...\cdot\frac{n+n}{n}$$ Then it becomes apparent that each term is always greater than one, and thus the sum obviously diverges.
Franklin Pezzuti Dyer
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Indeed the summands are $\binom{2n}{n}$, which we recognize as the maximum of all binomial coefficients choosing from $2n$. – hardmath Jun 27 '17 at 20:29
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@user458361 No problem, I'm glad to help.
If you found it helpful, don't forget to $\uparrow$ or $\checkmark$! :)
– Franklin Pezzuti Dyer Jun 27 '17 at 20:31
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$$\frac{(2n)!}{n!^2}=\frac{1\cdot 2\cdot\ldots \cdot n\,\cdot\,(n+1)\cdot\ldots \cdot 2n}{1\cdot 2\cdot\ldots \cdot n\,\cdot\,1\cdot 2\cdot\ldots \cdot n}$$ After cancellatoin, the numerator consists of $n$ factors $\ge n$, the denominator of $n$ factors $\le n$. Hence the general summand is $\ge 1$, but miust tend to $0$ for convergence.
Hagen von Eitzen
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