Krdytkyu

Can you help me to show that:

$$lim_{k to infty} frac{2^kk!}{sqrt{(2k+1)!}}=0$$

By cross-multiplication, for $kgt0$, we get
$$
left(frac{2k}{2k+1}right)^2ltfrac{2k}{2k+2}tag1
$$
Therefore,
$$
begin{align}
a_k^4
&=left(frac{color{#C00}{2^kk!},color{#090}{2^kk!}}{color{#090}{(2k+1)!}}right)^2tag2\
&=left(frac{color{#C00}{2cdot4cdots2k}}{color{#090}{3cdot5cdots(2k+1)}}right)^2tag3\
&ltfrac{2cdot4cdots2k}{4cdot6cdots(2k+2)}tag4\[3pt]
&=frac1{k+1}tag5
end{align}
$$
Explanation:
$(2)$: expand the terms to the fourth power
$(3)$: keep the red factors in the numerator and cancel the green factors in the denominator
$(4)$: apply $(1)$
$(5)$: cancel the factors in the numerator and denominator

For $k=5$, the square of the general term can be written

$$frac{2cdot2cdot4cdot4cdot6cdot6cdot8cdot8cdot10cdot10}{2cdot3cdot4cdot5cdot6cdot7cdot8cdot9cdot10cdot11}=
frac{2cdot4cdot6cdot8cdot10}{3cdot5cdot7cdot9cdot11}.$$

Taking the logarithm,

$$logfrac23+logfrac45+logfrac67+logfrac89+logfrac{10}{11}<-frac13-frac15-frac17-frac19-frac1{11}.$$

Generalizing to the $k^{th}$ term, the logarithm diverges to $-infty$.

An elementary way using

• 
  $(star)$: $prod_{i=1}^k(1+a_i)geq 1 + sum_{i=1}^k a_i$ for $a_i geq 0$ ($i=1,ldots , k$) and $k in mathbb{N}$
  

$$frac{2^kk!}{sqrt{(2k+1)!}}= frac{2^kk!}{sqrt{prod_{i=1}^k 2i cdot prod_{i=1}^k (2i+1)}} = frac{1}{sqrt{prod_{i=1}^k left(1+frac{1}{2i} right)}}$$
$$stackrel{(star)}{leq} frac{1}{sqrt{1+frac{1}{2}sum_{i=1}^k frac{1}{i}}}stackrel{k to infty}{longrightarrow}0$$

By Stirling approximation

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

$$frac{2^kk!}{sqrt{(2k+1)!}}=frac{2^ksqrt{2 pi k}left(frac{k}{e}right)^k}{sqrt[4]{2 pi (2k+1)}left(frac{2k+1}{e}right)^{k+frac12}}sim ccdotfrac{k^{k+frac12}}{k^{k+frac34}}=frac{c}{k^{frac14}}to 0$$