Using convergence/divergence tests on (2^[2k]*k!*k!)/(2k!)

[ksdhart]

I'm having difficulties with a problem from my Calc 3 course. The problem text says:

In exercises 29-34, use the ratio test to analyze whether the given series converges or diverges. If the ratio test is inconclusive, use a different test to analyze the series.

33. $\displaystyle \displaystyle \sum _{k=1}^{\infty }\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \left(2k\right)}{1\cdot 3\cdot 5\cdot \cdot \cdot \left(2k-1\right)}$

My first step was to convert the numerator and denominator into expressions of k-factorial and apply the ratio test:

$\displaystyle 2\cdot 4\cdot 6\cdot \cdot \cdot \left(2k\right)=2^k\left(1\cdot 2\cdot 3\cdot \cdot \cdot k\right)=2^k\cdot k!$

$\displaystyle 1\cdot 3\cdot 5\cdot \cdot \cdot \left(2k-1\right)=\frac{\left(2k\right)!}{2\cdot 4\cdot 6\cdot \cdot \cdot \left(2k\right)}=\frac{\left(2k!\right)}{2^k\cdot k!}$

$\displaystyle a_k=\left(2^k\cdot k!\right)\cdot \frac{2^k\cdot k!}{\left(2k\right)!}=\frac{2^{2k}\cdot k!\cdot k!}{\left(2k\right)!}$

$\displaystyle a_{k+1}=\frac{2^{2k+2}\cdot \left(k+1\right)!\cdot \left(k+1\right)!}{\left(2k+2\right)!}$

$\displaystyle \frac{a_{k+1}}{a_k}=\frac{2^{2k+2}\cdot \left(k+1\right)!\cdot \:\left(k+1\right)!}{\left(2k+2\right)!}\cdot \frac{\left(2k\right)!}{2^{2k}\cdot k!\cdot k!}=\frac{2^{2k+2}\cdot \left(k+1\right)!\cdot \left(k+1\right)!\cdot \left(2k\right)!}{\left(2k+2\right)!\cdot 2^{2k}\cdot k!\cdot k!}$

$\displaystyle \frac{a_{k+1}}{a_k}=\frac{4\left(k+1\right)\left(k+1\right)}{\left(2k+1\right)\left(2k+2\right)}= \frac{2k+2}{2k+1}$

With the expression in a simpler form, I took the limit:

$\displaystyle \displaystyle \rho =\lim _{k\to \infty }\left(\frac{a_{k+1}}{a_k}\right)=\lim _{k\to \infty }\left(\frac{2k+2}{2k+1}\right)=\lim _{k\to \infty }\left(\frac{2}{2}\right)=1$

Since rho is 1, the ratio test is inconclusive. Now, the instructions say to use a different test. But that's where I'm having problems. I thought using the root test might be my best bet.

$\displaystyle \displaystyle \lim _{k\to \infty }\left(\sqrt[k]{\frac{2^{2k}\cdot k!\cdot k!}{\left(2k\right)!}}\right)=\lim _{k\to \infty }\left(\frac{4\cdot \sqrt[k]{k!}\cdot \sqrt[k]{k!}}{\sqrt[k]{\left(2k\right)!}}\right)$

I know from a previous exercise that $\displaystyle \displaystyle \lim _{k\to \infty }\left(\sqrt[k]{k!}\right)=\infty$ and from that I also know that $\displaystyle \displaystyle \lim _{k\to \infty }\left(\sqrt[k]{\left(2k\right)!}\right)=\infty$. However, that still leaves me with an indeterminate form. I'd need to figure out whether the numerator grows faster than the denominator or vice versa. And I suspect I'd need to use L'Hôpital's rule for that... but I don't know how to take the derivative of k-factorial, so I'm at a loss right now.

Any help would be greatly appreciated.

 

[daon2]

Notice that for $\displaystyle n> 1$,

$\displaystyle a_n = \dfrac{2}{1}\cdot \dfrac{4}{3}\cdot \dfrac{6}{5} \cdots \dfrac{2n}{2n-1} > 2\cdot 1\cdot 1\cdots 1$

 

[ksdhart]

Wow. I was way overthinking this then. I got stuck in the mindset of using the closed form for the sequence, and I failed to see what was right in front of me. Each successive term in the sequence is more than 2, so the series must diverge. Thanks for the course correction.