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