# Prove convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{2\cdot 4\cdot 6\cdot\ldots\cdot 2n}{3\cdot 5\cdot7\cdot\ldots\cdot (2n+1)}$

Taken from Soo T. Tan's Calculus textbook Chapter 9.7 Exercise 27-

Define $$a_n=\frac{2\cdot 4\cdot 6\cdot\ldots\cdot 2n}{3\cdot 5\cdot7\cdot\ldots\cdot (2n+1)}$$ One needs to prove the convergence or divergence of the series $$\sum_{n=1}^{\infty} a_n$$

upon finding the radius of convergence for $\sum_{n=1}^{\infty}\frac{2\cdot 4\cdot 6\cdot\ldots\cdot 2n}{3\cdot 5\cdot7\cdot\ldots\cdot (2n+1)}\cdot x^{2n+1}$ to be $1$ and checking the endpoints. Also, please use tests and methods that are taught in introductory courses.

Answers show divergence but no without explanation.

• Do you mean the series $\sum_{n=1}^{\infty} a_n$ or the sequence $\{a_n\}_{n=1}^{\infty}$? – lulu Aug 27 '16 at 0:47
• Show that $a_n > \frac 1n$ when $n>2$ And the series diverges by the comparison test. – Doug M Aug 27 '16 at 0:50
• I mean series couldn't find the sigma button, sorry. @DougM Thanks – idanp Aug 27 '16 at 0:57
• @idanp I edited your post to include the summation symbol. If you are interested, click on the "edit" button to see the syntax I used. – lulu Aug 27 '16 at 0:59
• Thank you. @DougM is there a quick way to show it is larger than the harmonic series without induction? – idanp Aug 27 '16 at 1:06

Rewrite the $n$th term by sliding each of the factors in the numerator one position to the left. This gives $$a_n = \frac 21\frac43\frac65\cdots\frac{2n}{2n-1}\frac1{2n+1}.$$ We now see $a_n$ is a product consisting of factors bigger than one, multiplied onto the final factor $\frac1{2n+1}$. Conclude $$a_n>\frac1{2n+1},$$ so the series $\sum a_n$ diverges.

• Great, thanks a lot – idanp Aug 27 '16 at 1:34

Let

$$a = \frac{2}{3} \cdot \frac{4}{5} \cdots \frac{2n}{2n+1} , \quad b = \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n-1}{2n}$$

Then $a > b$ and $ab = \dfrac{1}{2n+1}$, so actually $a > \dfrac{1}{\sqrt{2n+1}}$ - stronger than what you needed.

In other words: You can use this to prove that even $\sum a_n^2$ diverges, which is stronger than the original question.

By using Euler's beta function we have $$\frac{(2n)!!}{(2n+1)!!} = \frac{4^n n!^2}{(2n+1)!} = 4^n B(n+1,n+1)$$ hence: $$\begin{eqnarray*} \sum_{n=1}^{N}\frac{(2n)!!}{(2n+1)!!} = \int_{0}^{1}\sum_{n=1}^{N}\left(4x(1-x)\right)^n\,dx&=&\int_{-1/2}^{1/2}\sum_{n=1}^{N}(1-4x^2)^n\,dx\\&=&\int_{0}^{1}\sum_{n=1}^{N}(1-x^2)^n\,dx\end{eqnarray*}$$ but the last integrand function, over the interval $(0,1)$, is bounded below by $$\max\left(0,N-\frac{N(N+1)}{2}x^2\right)$$ hence it follows that $$\sum_{n=1}^{N}\frac{(2n)!!}{(2n+1)!!} \geq \int_{0}^{\sqrt{\frac{2}{N+1}}}\left(N-\frac{N(N+1)}{2}x^2\right)\,dx = \frac{2\sqrt{2}\,N}{3\sqrt{N+1}}$$ and the original series is divergent. By telescoping we also have the identity $$\sum_{n=1}^{N}\frac{(2n)!!}{(2n+1)!!} = -2+\frac{\Gamma(N+2)}{\Gamma\left(N+\frac{3}{2}\right)}\sqrt{\pi}$$ and the improved bound $$\boxed{\; \sum_{n=1}^{N}\frac{(2n)!!}{(2n+1)!!} > -2+\sqrt{\pi(N+1)}\;}$$ follows from Gautschi's inequality.

Another way to look at the problem could be to consider that $$a_n=\frac {\prod_{i=1}^{n }(2i) }{\prod_{i=1}^{n }(2i+1) }$$ Using the ratio test $$\frac{a_{n+1}}{a_n}=\frac{2 (n+1)}{2 n+3}$$ which is inconclusive.

Using Raabe's test $$n\left(\frac{a_n}{a_{n+1}} -1\right)=\frac{n}{2 n+2}$$ its limit is $\frac 12$ and then divergence.

Another way, using

$$a_n=\frac {\prod_{i=1}^{n }(2i) }{\prod_{i=1}^{n }(2i+1) }=\frac {2^n\,n!}{\frac{2^{-n} (2 n+1)!}{n!}}=\frac{4^n\, (n!)^2}{(2n+1)! }$$ and using, for large values of $n$, Stirling approximation for the factorial leads to $$\log(a_n)=\left(-\frac{1}{2} \log (n)+\log \left(\frac{\sqrt{\pi }}{2}\right)\right)+O\left(\frac{1}{n}\right)$$ Using the fact that $a_n=e^{\log(a_n)}$, this leads to $$a_n=\frac{1}{2} \sqrt{\pi } \sqrt{\frac{1}{n}}+O\left(\frac{1}{n}\right)$$ which is divergent by comparaison to the $p$-series.