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I need to study the convergence of the series $\sum_{n=1}^{\infty}\frac{1^2*5^2*...*(4n-3)^2}{3^2*7^2*...*(4n-1)^2}$.

Now, I think we can do it by using the fact that if we have a series $\sum_{n=1}^{\infty}a_n$ and we can find $b_n$ so that $a_n<b_n$ then:

if $\sum_{n=1}^{\infty}b_n$ is convergent then $\sum_{n=1}^{\infty}a_n$ i convergent

or

if $\sum_{n=1}^{\infty}a_n$ is divergent then $\sum_{n=1}^{\infty}b_n$ is divergent

What I did so far is:

I said that $\frac{1}{3}<\frac{3}{5},\frac{5}{7}<\frac{7}{9},...,\frac{4n-3}{4n-1}<\frac{4n-1}{4n+1}$

and we get that

$\frac{1*3*...*(4n-3)}{3*7*...*(4n-1)}<\frac{3*7*...*(4n-1)}{5*9*...*(4n+1)}$

and if we multiply both sides with $\frac{1*3*...*(4n-3)}{3*7*...*(4n-1)}$

we get that $\frac{1^2*5^2*...*(4n-3)^2}{3^2*7^2*...*(4n-1)^2}<\frac{1}{4n-1}$ but that does not really help me.

uniquesolution
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Ghost
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  • Standard notation in mathematics does not require the symbol * to denote multiplication. Suggestion: use $x\cdot y$ instead of $x*y$. – uniquesolution Oct 27 '18 at 14:38
  • Two ideas : instead of considering product - consider the $e^{\log (\dots)}$ and try to complete numerator to factorial and complete denominator to product of odd numbers and product of even numbers. – openspace Oct 27 '18 at 14:40

1 Answers1

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We are interested in

$$ \sum_{n\geq 1}\left[\frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(n+\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4}\right)\Gamma\left(n+\frac{3}{4}\right)}\right]^2 $$ and by Gautschi's inequality the main term of this series behaves like $\frac{K}{n}$ as $n\to +\infty$, hence the given series is divergent by asymptotic comparison with the harmonic series.
Namely $K=\frac{1}{2\pi^2}\Gamma\left(\frac{3}{4}\right)^4$.

Elementary alternative: you may prove by induction that for any $n\in\mathbb{N}^+$ $$\left(\prod_{k=1}^{n}\frac{4k-3}{4k-1}\right)^2 \geq \frac{1}{9n} $$ holds (since $\left(\frac{4n+1}{4n+3}\right)^2-\frac{n}{n+1}=\frac{1}{(n+1)(4n+3)^2}>0$) and the conclusion is just the same.

Jack D'Aurizio
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  • This is a bit overkill if u ask me. I do not know the concepts that you used and it is a bit hard for me to understand. I am sure it can be done alot easier using basic concepts. – Ghost Oct 27 '18 at 15:03
  • @Ghost: I added an elementary derivation. – Jack D'Aurizio Oct 27 '18 at 15:06
  • Ok, I understand that you can prove by induction that inequality but can you explain how you got that inequality? – Ghost Oct 27 '18 at 15:08
  • @Ghost: not by chance $K$ is close to (greater than) $\frac{1}{9}$, so the second inequality stood a high probability to hold in advance. Indeed it holds since $\left(\frac{4n+1}{4n+3}\right)^2\geq \frac{n}{n+1}$. – Jack D'Aurizio Oct 27 '18 at 15:10