Find if this series is convergent or divergent

Question

So I have been given this question by our professor as an extra question and I have solved it to some extent but I've been stuck, I'll write my approach in here and I would like some help. So this is the series I have been given: $$\sum_{n=1}^\infty \left(\frac{1\times3\times ...\times(2n-1)}{2\times4\times...\times2n}\right)^P$$ I used the ratio test to prove that this series is either divergent or convergent: $$\frac{a_{n+1}}{a_n}=\left(\frac{\frac{1\times3\times...(2n-1)\times(2n+1)}{2\times4\times...\times(2n)\times(2n+2)}}{\frac{1\times3\times...\times(2n-1)}{2\times4\times...\times2n}}\right)^P$$ Which is equal to: $$\frac{a_{n+1}}{a_n}=\left(\frac{2n+1}{2n+2}\right)^P$$ Now if we take the limit of this we would have: $$\lim_{n\rightarrow\infty} \left(\frac{2n+1}{2n+2}\right)^P = \lim_{n\rightarrow\infty}\left(\frac{2+\frac{1}{n}}{2+\frac{2}{n}}\right)^P = \left(\frac{2}{2}\right)^P = 1^P $$ Which for every P we get 1.

So the ratio test fails to determine either this series is convergent of divergent. I have done some researching and there is a test called Sterling's approximation but we haven't studied that yet and I have to comment that this question came up in my real analysis course so if there is another "analytic" way to solve this I would appreciate the help.

Answer 1

The Frink's convergence test can be used here.

Let $\displaystyle a_n(p)=\left(\frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot\ldots\cdot(2n)}\right)^p$. Then

$$\Big(\frac{a_{n+1}(p)}{a_n(p)}\Big)^n=\Big(\frac{2n+1}{2n+2}\Big)^{np}=\frac{1}{\left(1+\frac{1}{2n+1}\right)^{np}}=\frac{1}{\left(1+\tfrac{np\tfrac{1}{2n+1}}{np}\right)^{np}}\xrightarrow{n\rightarrow\infty}\frac{1}{e^{p/2}}$$

• For $p>2$, $\displaystyle \frac{1}{e^{p/2}}<\frac1e$ and so, $\sum_na_n(p)$ converges.
• If $0<p<2$, $\displaystyle \frac{1}{e^{p/2}}>\frac1e$ and so, $\sum_na_n(p)$ diverges to $\infty$.
• For the case $p=2$ notice that $\Big(1+\frac1{2n+1}\Big)^{2n}<\Big(1+\frac1{2n}\Big)^{2n}\nearrow e$ and so, $$\frac{1}{\big(1+\frac1{2n+1}\big)^{2n}}>\frac1e, \quad n\geq 1$$ By Frink's test, $\sum_na_n(2)$ diverges to $\infty$.