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The following series $$\sum_{n=0}^{\infty} \frac{(n!)^{2}}{(2n)!}$$ converges. It fails the divergence test, but once I apply the ratio test, the limit is always equal to $\infty$. Unless you cannot distribute an exponent on a factorial product, I don't see what I am doing wrong.

I tried using Stirling's approximation to construct a geometric series, but I always got $1$ for the ratio, so the test remains inconclusive.

As for the other tests, I didn't find any similar series against which to compare this one to (the limit comparison always gave me $\infty$ or $0$...).

Any help or tips on how to take a different spin on the problem are greatly appreciated.

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    $\begingroup$ Fun fact: it converges to $\frac{4}{3}+\frac{2 \pi }{9 \sqrt{3}}$. $\endgroup$
    – Winther
    Jan 10, 2016 at 14:52
  • $\begingroup$ @Winther Sure sounds like fun! How did you compute this? $\endgroup$ Jan 10, 2016 at 15:18
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    $\begingroup$ We can use the power-series for $\arcsin^2(x)$ to show that $\sum_ {n = 0}^\infty \frac {x^{2 n}} {{2 n\choose n}} = \frac{2 x \arcsin\left(\frac{x}{2}\right)}{\left(4-x^2\right)^{3/2}}+\frac{4}{4-x^2}$ from which it follows. See e.g. this question and refs. within. $\endgroup$
    – Winther
    Jan 10, 2016 at 15:20
  • $\begingroup$ Oh wow, power series suddenly just got relevant...That's awesome though! $\endgroup$ Jan 10, 2016 at 15:23
  • $\begingroup$ Formula, closed form are in oeis.org/A091682 . $\endgroup$ Mar 19 at 15:38

8 Answers 8

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$$\begin{eqnarray*}\sum_{n\geq 0}\frac{n!^2}{(2n)!}=\sum_{n\geq 0}\frac{\Gamma(n+1)^2}{\Gamma(2n+1)}&=&\sum_{n\geq 0}(2n+1)\int_{0}^{1}x^n(1-x)^n\,dx\\&=&\int_{0}^{1}\frac{1+x-x^2}{(1-x+x^2)^2}\,dx\\&=&\left.\frac{2}{3}\cdot\frac{2x-1}{1-x+x^2}+\frac{2}{3\sqrt{3}}\cdot\arctan\left(\frac{2x-1}{\sqrt{3}}\right)\right|_{0}^{1}\\&=&\color{red}{\frac{2}{27}(18+\pi\sqrt{3}).}\end{eqnarray*}$$

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  • $\begingroup$ Neat evaluation! $\endgroup$ Jan 12, 2019 at 17:27
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We have the following series:

$$\sum_{n=0}^{\infty} \frac{(n!)^{2}}{(2n)!}$$

To prove convergence, use the Ratio Test:

$$\lim_{n\to\infty}\left|\frac{\left((n+1)!\right)^{2}}{(2(n+1))!}*\frac{(2n)!}{(n!)^{2}}\right|$$ $$\lim_{n\to\infty}\left|\frac{(n+1)!(n+1)!}{(2n+2)!}*\frac{(2n)!}{(n!)(n!)}\right|$$ $$\lim_{n\to\infty}\left|\frac{(n+1)(n+1)}{(2n+1)(2n+2)}\right|$$

As you notice, the leading degree of the denominator is the same as the leading degree of the numerator. Therefore, we look at the coefficients, and so the limit goes to $\frac{1}{4}$, and the series convergences absolutely by the Ratio Test.

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    $\begingroup$ Yeah I kept forgetting that $(2n + 2)!$ translated to $(2n + 2)(2n + 1)(n!)$ so the denominator was always greater by one degree...I feel silly now, thanks for writing all that out! $\endgroup$ Jan 10, 2016 at 14:58
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    $\begingroup$ @AndresStadelmann np. Always happy to help! $\endgroup$
    – Varun Iyer
    Jan 10, 2016 at 14:59
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By Stirling:

$$\frac{(n!)^2}{(2 n)!} \sim \frac{n^{2 n} e^{-2 n} 2 \pi n}{(2 n)^{2 n} e^{-2 n} \sqrt{2 \pi 2 n}} = \frac{\sqrt{2 \pi n}}{\sqrt{2} 2^{2 n}}$$

so the sum certainly converges by comparison with

$$\sum_{n=0}^{\infty} \sqrt{n} \, 2^{-2 n} $$

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  • $\begingroup$ Isn't that second series that you're using for comparison smaller than the first one? $\endgroup$ Jan 10, 2016 at 14:50
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    $\begingroup$ @AndresStadelmann Are you referring to the last line? (if so, the difference is only by a constant factor, namely dropping the $\pi$) $\endgroup$
    – Clement C.
    Jan 10, 2016 at 15:02
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    $\begingroup$ @AndresStadelmann: "by comparison with" means that, because the asymptotic behavior of each term in series A matches that in series B, and we know series A converges, then series B converges. $\endgroup$
    – Ron Gordon
    Jan 10, 2016 at 15:07
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    $\begingroup$ You can add the $\pi$ back, but since it is a constant it does not really matter: for any constant $c\neq 0$, $\sum_n a_n$ converges iff $\sum_n c a_n$ converges. $\endgroup$
    – Clement C.
    Jan 10, 2016 at 15:07
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    $\begingroup$ @AndresStadelmann To keep a small caveat to the asymptotics method: "if $a_n \sim_{n\to\infty} b_n$, then $\sum_n a_n$ and $\sum_n b_n$ have same nature" only holds if (at least one of) $a_n$ or $b_n$ keeps a constant sign (either non-negative or non-positive) for $n$ big enough. Otherwise, it may fail: consider e.g. $a_n = \frac{(-1)^n}{\sqrt{n}}$ and $b_n = a_n + \frac{1}{n}$. $\endgroup$
    – Clement C.
    Jan 10, 2016 at 16:10
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If you apply the ratio test you should get the limit of a(n+1)/a(n) tends to 1/4 hence we get convergence.

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  • $\begingroup$ Of course it was only an algebraic mistake, why would it have been anything else...I feel kinda dumb $\endgroup$ Jan 10, 2016 at 14:59
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The ratio test gives you $$\frac{(2n)!}{(n!)^2} \cdot \frac{((n+1)!)^2}{(2n+2)!} = \frac{(n+1)(n+1)}{(2n+2)(2n+1)} \to \frac 14 < 1$$ as $n$ tends to infinity. So the series is convergent.

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  • $\begingroup$ I'm pretty sure you inverted that third fraction, but yeah, I realized I was forgetting the extra $(2n + 1)$ so the ratio was always $\infty$. Thanks! $\endgroup$ Jan 10, 2016 at 14:55
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    $\begingroup$ Sorry for the typo ;) $\endgroup$
    – Crostul
    Jan 10, 2016 at 14:56
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You can use the root test.

$$\sqrt[n]{\frac{(n!)^2}{(2n)!}}=\sqrt[n]{\frac{2\cdot 3\cdots n}{(n+1)\cdots(2n)}}\leq \frac{1}{n}\sum_{j=1}^n \frac{j}{n+j}\leq \frac{1}{n} \sum_{j=1}^{n}\frac{j}{j+j}=\frac{1}{2},$$(by AM-GM inequality) so$$\limsup_{n\to \infty} \sqrt[n]{\frac{(n!)^2}{(2n)!}} \leq \frac{1}{2}.$$ Therefore the series converges.

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  • $\begingroup$ Does this also go by the name of the "Cauchy criterion" or something like that? $\endgroup$ Jan 16, 2016 at 8:12
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    $\begingroup$ As far as I know "Cauchy criterion" refers to "checking wether the sequence of partial sums is Cauchy". But you're right, it is also called "Cauchy root test" according to Wikipedia en.wikipedia.org/wiki/Root_test. $\endgroup$
    – Arnaud D.
    Jan 16, 2016 at 10:34
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You can conclude by comparison, since $$\frac{(n!)^2}{(2n)!} \sim_{n\to\infty} \frac{\sqrt{\pi n}}{2^{2n}}$$ by Stirling's approximation.

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    $\begingroup$ Because I know this one by heart, and it's very easy to invoke comparison theorems afterwards. Did you downvote because you think it's overkill? $\endgroup$
    – Clement C.
    Jan 10, 2016 at 14:47
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    $\begingroup$ To be more specific: $\frac{1}{2^n}\binom{n}{n/2}$ is the probability to get a $n$-bit string in the middle layer of the Boolean hypercube, under the uniform distribution over $\{0,1\}^n$. This tends to arise quite a lot in computer science, so the asymptotics are a useful thing to know... and once memorized, it's not that "dumb" to apply it. $\endgroup$
    – Clement C.
    Jan 10, 2016 at 14:50
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    $\begingroup$ The proof of Stirling's formula is quite complicated and I have seen it only after a lot of complex analysis. On the other hand, I was able to detect the convergence of this series simply after studying basic calculus. Since the OP said that he failed using the ratio test, I thought it would have been more instructive to use basic tools to solve the problem. $\endgroup$
    – Crostul
    Jan 10, 2016 at 14:55
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    $\begingroup$ Well, it depends on what you consider being basics: I recall learning a proof of Stirling's in high school, so by now I consider it as part of the toolbox (if you don't care about the exact constant in the equivalent, then there are more elementary proofs.); and a tool definitely worth knowing. Moreover, the question does state "Any help or tips on how to take a different spin on the problem are greatly appreciated." $\endgroup$
    – Clement C.
    Jan 10, 2016 at 15:01
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    $\begingroup$ @Crostul: not for nothing, the OP stated that he tried Stirling and got nowhere so I thought that's how he wanted to see the problem solved. Perhaps it is the idea of not reading the problem thoroughly that is "dumb." $\endgroup$
    – Ron Gordon
    Jan 10, 2016 at 15:06
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For $n≥1$, easy to know: $\frac{n!^2}{(2n)!}\le\frac{1}{2^n}$, hence the answer is obvious.

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