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I am trying to determine whether or not the following series converges$$\sum_{n=1}^\infty \frac{1*3*5*...*(2n+1)}{2*5*8*...*(3n+2)}$$My first attempt at figuring this out involved the fact that I know $$\prod_{i=1}^n(2i-1)=\frac{(2n)!}{2^nn!}$$It seems like I could simply change this to $\frac{(2(n+1))!}{2^{n+1}(n+1)!}$ and it would work in this scenario. However, I'm not certain of that, and even if it did work I have no idea how to approach the denominator. What am I missing?

Andrew
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3 Answers3

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$\sum_\limits{n=1}^\infty \frac{1*3*5*...*(2n+1)}{2*5*8*...*(3n+2)}$

Ratio test:

$\lim_\limits {n\to \infty}\frac {a_{n+1}}{a_n} = \lim_\limits {n\to \infty}\frac {2n+3}{3n+5} = \frac 23 <1$

The series converges.

Doug M
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We have: $$ S=\sum_{n\geq 1}\prod_{k=0}^{n}\frac{2k+1}{3k+2}=\frac{\Gamma\left(\frac{5}{3}\right)}{\sqrt{\pi}}\sum_{n\geq 1}\frac{\Gamma\left(n+\frac{3}{2}\right)}{\Gamma\left(n+\frac{5}{3}\right)}\left(\frac{2}{3}\right)^n$$ hence the series is convergent by comparison with a convergent geometric series, due to $\Gamma\left(n+\frac{3}{2}\right)<\Gamma\left(n+\frac{5}{3}\right)$. Additionally, by Gautschi's inequality we have

$$ S < \frac{\Gamma\left(\frac{5}{3}\right)}{\sqrt{\pi}}\sum_{n\geq 1}\frac{\left(\frac{2}{3}\right)^n}{\left(n+\frac{2}{3}\right)^{1/6}} $$

and by exploiting Euler's beta function we have $$\begin{eqnarray*} S &=& \frac{\Gamma\left(\frac{5}{3}\right)}{\sqrt{\pi}\,\Gamma\left(\frac{1}{6}\right)}\int_{0}^{1}\sum_{n\geq 1}\left(\frac{2}{3}\right)^n x^{n+\frac{1}{2}}(1-x)^{-\frac{5}{6}}\,dx\\ &=& \color{red}{\frac{2\,\Gamma\left(\frac{5}{3}\right)}{\sqrt{\pi}\,\Gamma\left(\frac{1}{6}\right)}\int_{0}^{1}\frac{x^{3/2}}{(1-x)^{1/6}(3-2x)}\,dx}. \end{eqnarray*}$$

Community
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Jack D'Aurizio
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  • Why the downvote? – Jack D'Aurizio Sep 14 '16 at 23:09
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    +1 for multiple methods. I started considering a ratio of Gamma functions, but saw that the ratio was less than $\left(\frac23\right)^{n+1}$. This hinted to me that there might be a simpler way. It might be that the downvoter didn't like the complexity of your methods (however, that seems to me to possibly inspire a curious person to investigate further). – robjohn Sep 14 '16 at 23:38
  • Note that $\Gamma!\left(n+\frac32\right)\le\Gamma!\left(n+\frac53\right)$. Thus, Gautschi's Inequality is not really needed. We can use that $\Gamma'!\left(\frac32\right)=\sqrt\pi\left(1-\log(2)-\frac\gamma2\right)\gt0$ to show the inequality for $n=0$. – robjohn Sep 15 '16 at 13:39
  • @robjohn: answer slightly modified to clarify what I meant before. – Jack D'Aurizio Sep 15 '16 at 13:46
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Hint:

Since $\frac{2k+1}{3k+2}\le\frac23$ $$ \begin{align} \frac{1*3*5*...*(2n+1)}{2*5*8*...*(3n+2)} &\le\left(\frac23\right)^{n+1} \end{align} $$

robjohn
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