5

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.

Andres Stadelmann
  • 151

8 Answers8

12

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

Jack D'Aurizio
  • 353,855
9

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.

Varun Iyer
  • 6,074
5

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

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

Ron Gordon
  • 138,521
  • Isn't that second series that you're using for comparison smaller than the first one? – Andres Stadelmann Jan 10 '16 at 14:50
  • 1
    @AndresStadelmann Are you referring to the last line? (if so, the difference is only by a constant factor, namely dropping the $\pi$) – Clement C. Jan 10 '16 at 15:02
  • @ClementC. Yeah so shouldn't that make all the terms in the series with $\pi$ slightly bigger? – Andres Stadelmann Jan 10 '16 at 15:06
  • 1
    @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. – Ron Gordon Jan 10 '16 at 15:07
  • 1
    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. – Clement C. Jan 10 '16 at 15:07
  • Again, my algebra is super sloppy: I was cancelling out the $2^{2n}$ every time -_- – Andres Stadelmann Jan 10 '16 at 15:10
  • @RonGordon Interesting, I've only ever used the comparison test with series greater or larger than the one in question, but apparently "equal to" works as well. I'll keep that in mind for future problems, thanks! – Andres Stadelmann Jan 10 '16 at 15:13
  • @ClementC. Awesome, good to know – Andres Stadelmann Jan 10 '16 at 15:14
  • 2
    @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 $b_n = a_n + \frac{1}{n}$. – Clement C. Jan 10 '16 at 16:10
5

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.

M.crolla
  • 125
3

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.

Crostul
  • 36,738
  • 4
  • 36
  • 72
3

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.

Arnaud D.
  • 20,884
  • Does this also go by the name of the "Cauchy criterion" or something like that? – Andres Stadelmann Jan 16 '16 at 08:12
  • 1
    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 https://en.wikipedia.org/wiki/Root_test. – Arnaud D. Jan 16 '16 at 10:34
2

You can conclude by comparison, since $$\frac{(n!)^2}{(2n)!} \sim_{n\to\infty} \frac{\sqrt{\pi n}}{2^{2n}}$$ by Stirling's approximation.

Clement C.
  • 67,323
  • 1
    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? – Clement C. Jan 10 '16 at 14:47
  • 1
    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. – Clement C. Jan 10 '16 at 14:50
  • 1
    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. – Crostul Jan 10 '16 at 14:55
  • 2
    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." – Clement C. Jan 10 '16 at 15:01
  • @Crostful You're probably right, and since Stirling was never formally introduced in our calculus class, there shouldn't really be a need to use it for any of our exercises. That being said, an extra way to look at a problem never hurts, and it has paid off for me in the past, so it's good to keep it handy (even though I have no clue what the proof is, this is the case for most of the theorems I use in the first place...). – Andres Stadelmann Jan 10 '16 at 15:03
  • 1
    @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." – Ron Gordon Jan 10 '16 at 15:06
1

For $n≥1$, easy to know: $\frac{n!^2}{(2n)!}\le\frac{1}{2^n}$, hence the answer is obvious.

Kai
  • 116