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$$\sum_{n=1}^\infty\frac{(2n)!}{2^{2n}(n!)^2}$$

Can I have a hint for whether this series converges or diverges using the comparison tests (direct and limit) or the integral test or the ratio test?

I tried using the ratio test but it failed because I got 1 as the ratio. The integral test seems impossible to use here.

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  • $\begingroup$ Your sum does not converge! $\endgroup$ Oct 19, 2018 at 12:54
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    $\begingroup$ @Dr.SonnhardGraubner That doesn't seem to be a very helpful hint or answer... $\endgroup$
    – DonAntonio
    Oct 19, 2018 at 13:00
  • $\begingroup$ Hint: Use Stirling's approximation on the factorials $\endgroup$ Oct 19, 2018 at 13:10
  • $\begingroup$ Ronald Blaak this was what I was discussing with my friends but since we haven't been taught it, I doubt we are allowed to use it. $\endgroup$ Oct 19, 2018 at 13:24

3 Answers 3

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For a more direct approach, you might directly expand the terms as follows: $$\begin{align} \frac{(2n)!}{4^n (n!)^2} &= \frac{1}{4^n}\frac{2n(2n-1)}{n^2}\frac{(2n-2)(2n-3)}{(n-1)^2}\cdots\frac{(4)(3)}{2^2} \frac{(2)(1)}{1^2} \\ &= \frac{2^n}{4^n}\frac{2n-1}{n} \frac{2n-3}{n-1}\cdots\frac{3}{2} \frac{1}{1} \\ &= \frac{4^n}{4^n} \frac{n - 1/2}{n} \frac{n - 3/2}{n-1} \cdots \frac{3/2}{2} \frac{1/2}{1}. \end{align}$$ This is almost a telescoping product. By subtracting $1/2$ from each numerator (except the last), we get a smaller term that does telescope. Thus $$ \frac{(2n)!}{4^n (n!)^2} \geq \frac{n-1}{n}\frac{n-2}{n-1} \cdots \frac{1}{2} \cdot (1/2) = \frac{1}{2n}. $$ Thus the $n$th term of your series is bigger than $1/2n$, and diverges by comparison with the harmonic series $$ \sum_{n \geq 1} \frac{1}{n}.$$

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    $\begingroup$ As a side comment --- when I checked this for myself, I also used Stirling's formula. But I know that I didn't know Stirling's bound when I first learned about series tests, so I understand the want for an answer without machinery. $\endgroup$
    – davidlowryduda
    Oct 19, 2018 at 13:55
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    $\begingroup$ Very nice answer to a rather difficult (undergraduate level without Stirling's Formula) question. +1 $\endgroup$
    – DonAntonio
    Oct 19, 2018 at 14:17
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Every term of the series is $$\frac{(2n)!}{2^{2n}(n!)^2}=\dfrac{\Gamma(2n+1)}{2^{2n}\Gamma^2(n)}=\dfrac{\Gamma(n+\frac12)}{n\Gamma(n)\sqrt{\pi}}>\dfrac{1}{2n}$$ by the formula $$\dfrac{\Gamma(n)}{\Gamma(2n)}=\dfrac{\sqrt{\pi}}{2^{2n-1}\Gamma(n+\frac12)}$$

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    $\begingroup$ Note that the Gamma function is increasing. $\endgroup$
    – Nosrati
    Oct 19, 2018 at 13:32
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    $\begingroup$ the moderators vote each other and leave people un-voted, it is ridiculous :) $\endgroup$
    – Nosrati
    Oct 19, 2018 at 14:00
  • $\begingroup$ I can't say what the moderators do, and I don't really care, though I think they are not a closed sect and "vote for each other". Your answer uses Gamma Function and some of its properties, which seems to be (usually) stuff from middle undergraduate level and up. Perhaps that's why your answer hasn't been upvoted yet. $\endgroup$
    – DonAntonio
    Oct 19, 2018 at 14:21
  • $\begingroup$ Lol. Btw you don't need votes, your answers are excellent. $\endgroup$
    – manooooh
    Oct 19, 2018 at 14:21
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HINT

We have that

$$\frac{(2n)!}{2^{2n}(n!)^2} =\frac1{4^n}\binom{2n}{n} \sim \frac{1}{\sqrt{2\pi n}}$$

indeed recall that by bounds and asymptotic formulas for the binomial coefficient

$$\binom{2n}{n} \sim \frac{4^n}{\sqrt{2\pi n}}$$

then we can refer to limit comparison test.

Refer also to the related

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