Infinite sum of products on my mid-term

Question

This problem was on my calculus mid-term : Determine the convergence or divergence of the series $$ \sum\limits_{n = 1}^\infty {\prod\limits_{k = 1}^n {\frac{{4k - 3}}{{4k - 1}}} } $$ I tried everything, even the Wallis product formula, but nothing worked out. I'm pretty sure it diverges but I can't find a way to prove that the general term doesn't tend to 0. The following question was (I'm only including this because it involves roughly the same series) :

Determine the convergence or divergence of the alternating series: $$ \sum\limits_{n = 1}^\infty {( - 1)^n \prod\limits_{k = 1}^n {\frac{{4k - 3}}{{4k - 1}}} } $$

Answer 1

Set $$a_n:=\prod_{k=1}^n\frac{4k-3}{4k-1}.$$ Now $$a_n\ge\frac{1}{3}\prod_{k=2}^n\frac{4k-5}{4k-1}=\frac{1}{4n-1},$$ which means that the first series diverges.

For the second series, consider $\log a_n$: $$\log a_n=\sum_{k=1}^n\log\left(1-\frac{2}{4k-1}\right) \le-\sum_{k=1}^n\frac{2}{4k-1}\to-\infty$$ as $n\to\infty$. This means $a_n\to 0$ as $n\to\infty$, and the alternating series converges relatively but (as we've seen from the first series) not absolutely.

Answer 2

$\prod\limits_{k = 1}^n {\frac{{4k - 3}}{{4k - 1}}}$ is decreasing.

Also if you note $u_n=\prod\limits_{k = 1}^n {\frac{{4k - 3}}{{4k - 1}}}$, you have $$\frac{u_{n+1}}{u_n}=1-\frac{2}{4n+3} \le 1-\frac{1}{2}\frac{1}{n+1} \le \frac{n^{1/4}}{(n+1)^{1/4}}$$

Consequently $u_n \le \frac{u_1}{n^{1/4}}=\frac{1}{3}\frac{1}{n^{1/4}}$ so $u_n$ converges to $0$ and according to the alternating series test the second series is convergent.

Answer 3

Consider $$S_m=\sum\limits_{n = 1}^m {\prod\limits_{k = 1}^n {\frac{{4k - 3}}{{4k - 1}}} }=\sum\limits_{n = 1}^m u_n$$ with $$u_n=\prod\limits_{k = 1}^n {\frac{{4k - 3}}{{4k - 1}}} $$ So $$\frac{u_{n+1}}{u_n}=\frac{(4n+1)}{(4n+3)}=1-\frac{2}{4 n+3}$$

I am sure that you can you take from here.

Answer 4

First use gamma function to write the product conveniently: $$\prod\limits_{k = 1}^n {\frac{{4k - 3}}{{4k - 1}}}=\frac{\Gamma(\frac{3}{4})\Gamma(m+\frac{1}{4})}{\Gamma(\frac{1}{4})\Gamma(m+\frac{3}{4})}$$ (factor $4^n$ out and the reshuffle the rest.)

Now note that $$\begin{align}\frac{m\Gamma(m+\frac{1}{4})}{\Gamma(m+\frac{3}{4})}&=\frac{m\Gamma(m+\frac{1}{4})}{(m-\frac{3}{4})\Gamma(m-\frac{3}{4})}\\ &>1, \end{align}$$ which implies that $$\frac{\Gamma(m+\frac{1}{4})}{\Gamma(m+\frac{3}{4})}>\frac{1}{m}.$$ Therefore your first series diverges.

Note that by the alternating series test and the fact that $\frac{\Gamma(m+\frac{1}{4})}{\Gamma(m+\frac{3}{4})}=O(m^{(-1/2)})$ the other series converges.