I know $4^{n}$ diverges, and $\frac{(n!)^2}{(2n)!}$ converges. I also think that their product diverges since the term of this series is increasing but I don't know how to prove it. I tried ratio test which is inconclusive, and some other tests.


marked as duplicate by Arnaud D., RRL, Xander Henderson, Namaste, mrtaurho Jan 13 at 21:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Use Stirling's approximation for the factorial term $\endgroup$ – asdf Apr 15 '18 at 16:33
  • 3
    $\begingroup$ It has been proved multiple times on MSE that $$\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}},$$ so... (you do not need Stirling's inequality/approximation to prove it) $\endgroup$ – Jack D'Aurizio Apr 15 '18 at 16:33
  • $\begingroup$ Wait - do you really ask for the series $\sum 4^n/{2n\choose n}$ or just for the sequence $\{4^n/{2n\choose n}\}_{n=0}^\infty$? $\endgroup$ – Hagen von Eitzen Apr 15 '18 at 16:43
  • $\begingroup$ I am asking for series. $\endgroup$ – Nebeski Apr 15 '18 at 16:47
  • $\begingroup$ More methods at math.stackexchange.com/questions/1606836/… $\endgroup$ – BAYMAX Apr 15 '18 at 17:17

Let $a_n:=4^n\cdot\frac{(n!)^2}{(2n)!}$, and note that $2^n\cdot n!=2\cdot4\cdot6\cdots 2n$. Hence

$$a_n=4^n \frac{(n!)^2}{(2n)!}=\prod_{k=1}^{n}\frac{2k}{2k-1}=\exp\left(\sum_{k=1}^n\log\left(\frac{2k}{2k-1}\right)\right)\ge \exp\left(n\log\left(\frac{2n}{2n-1}\right)\right)\tag1$$



Thus $\lim a_n\ge \sqrt e$, so $\sum a_n=\infty$.

A more simple approach is expanding and chopping the product on $(1)$, that is


Then we have the lower bound

$$a_n\ge 1+\sum_{k=1}^n\frac1{2k-1}=1+\frac12\sum_{k=1}^n\frac1{k-1/2}\ge 1+\frac12\sum_{k=1}^n\frac1k\tag4$$

Hence $\lim a_n=\infty$ (because $\sum\frac1k=\infty$).

  • $\begingroup$ Where does the exponent n disappear in your note? $\endgroup$ – Nebeski Apr 15 '18 at 17:17
  • $\begingroup$ I wasn't talking about that. $2^n×n!=2×4×\dots×2n$ only if exponent n=1 $\endgroup$ – Nebeski Apr 15 '18 at 17:24
  • $\begingroup$ I mean $2^n×n! = (2^n)×(2×2^n)×\dots×(n×2^n)$ $\endgroup$ – Nebeski Apr 15 '18 at 17:35
  • $\begingroup$ @Nebeski this $2^n×n! \neq (2^n)×(2×2^n)×\dots×(n×2^n)$. In the RHS you are multiplying by $2^n$ many times, and in LHS only once. $\endgroup$ – Masacroso Apr 15 '18 at 17:38
  • $\begingroup$ Ohhh yeah, I think all these $×$ confused me. $\endgroup$ – Nebeski Apr 15 '18 at 17:45

Stirling's approximation? Naah, let us go for a greater overkill. Since the series $$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^3 $$ is convergent to $\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}$ due to the relation with the squared complete elliptic integral of the first kind (identity $(7)$ at $k=\frac{1}{\sqrt{2}}$), its main term is convergent to zero and your sequence is divergent.

Seriously, an elementary approach. Since $2\cos\theta=e^{i\theta}+e^{-i\theta}$ and $\int_{0}^{2\pi}e^{ni\theta}e^{-mi\theta}\,d\theta = 2\pi\delta(m,n)$, $$ \frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}\theta\,d\theta. $$ By the dominated/monotone convergence theorem, the limit of both sides as $n\to +\infty$ is zero, hence your sequence is divergent. We also have that $\left\{\frac{1}{4^n}\binom{2n}{n}\right\}_{n\geq 1}$ is log-convex due to the Cauchy-Schwarz inequality and the previous integral representation.

Yet another elementary approach. You may prove in a combinatorial fashion that $$ \sum_{k=0}^{n}\binom{2k}{k}\binom{2n-2k}{n-k} = 4^n \tag{Convolution}$$ hence it follows that $$ \left[\sum_{k=0}^{n}\frac{1}{4^k}\binom{2k}{k}\right]^2 \leq \sum_{k=0}^{2n} 1 = 2n $$ and $$ \sum_{k=0}^{n}\frac{1}{4^k}\binom{2k}{k} \leq \sqrt{2n}. $$ On the other hand we also have

$$ \left[\sum_{k=0}^{n}\frac{1}{4^k}\binom{2k}{k}\right]^2 \geq \sum_{k=0}^{n} 1 = n $$ hence $$ \sqrt{n}\leq \sum_{k=0}^{n}\frac{1}{4^k}\binom{2k}{k} \leq \sqrt{2n}.\tag{SumInequality} $$ Since the sequence $\left\{\frac{1}{4^n}\binom{2n}{n}\right\}_{n\geq 1}$ is decreasing, the previous inequality implies that $\frac{1}{4^n}\binom{2n}{n}\ll\frac{1}{\sqrt{n}}$ as $n\to +\infty$. The correct asymptotic behaviour is given by Wallis' product and it is $$ \frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}.\tag{GoodToKnow}$$

  • 1
    $\begingroup$ Thanks for the "elementary" approach too! Stirling is very brutal and overkill but it works so well :) $\endgroup$ – gimusi Apr 15 '18 at 20:12

By Stirling's approximation

$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

we have

$$4^n \frac{(n!)^2}{(2n)!}\sim 4^n \frac{2 \pi n\left(\frac{n}{e}\right)^{2n}}{\sqrt{4 \pi n}\left(\frac{2n}{e}\right)^{2n}}=\sqrt{\pi n}\to \infty$$

then the series diverges.


Not the answer you're looking for? Browse other questions tagged or ask your own question.