Analyzing the series (2k)!/(k!)!

[ksdhart]

I've solved this problem in a very obtuse way, but I want to know if I'm missing something super obvious and there's an easier solution. I'm on quite a roll with this chapter as far as missing the obvious. The problem states:

For exercise #41-59, determine if the series converges or diverges

51. \(\displaystyle \displaystyle \sum _{k=0}^{\infty }\frac{\left(2k\right)!}{\left(k!\right)!}\)

My first step was to let \(\displaystyle a_k=\frac{\left(2k\right)!}{\left(k!\right)!}\). I then did the following:

\(\displaystyle \displaystyle ln\left(k!\right)=\sum _{x=1}^kln\left(x\right)\)

\(\displaystyle \displaystyle ln\left(k!\right)\ge \int _1^k\:ln\left(x\right)dx\)

\(\displaystyle ln\left(k!\right)\ge k\cdot ln\left(k\right)-k\)

\(\displaystyle k!\ge e^{k\cdot \:ln\left(k\right)}\cdot e^{-k}\)

\(\displaystyle k!\ge \frac{k^k}{e^k}\)

And from that, I have:

\(\displaystyle \frac{\left(2k\right)!}{\left(k!\right)!}\ge \frac{2k^{2k}}{e^{2k}}\cdot \frac{e^{k!}}{k!^{k!}}\)

So, I let \(\displaystyle b_k=\frac{2k^{2k}}{e^{2k}}\cdot \frac{e^{k!}}{k!^{k!}}\)

I have two series, such that a_k > b_k > 0 for all k >1 so I used the limit comparison test. But I first needed to prove that b_k diverges, so I used the root test on it.

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

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

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

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

The value of the second limit doesn't matter. Even if it goes to infinity, infinity times infinity is not an indeterminate form. Thus, by the root test, the series of b_k diverges. And since the series of a_k is greater than a divergent series, it too must diverge.

As I said above, my solution works... but I highly doubt that it's the intended solution. Can anyone provide insight into what I might have missed?

 

[lookagain]

\(\displaystyle \displaystyle =\infty \cdot \lim _{k\to \infty }\left(\sqrt[k]{\frac{e^{k!}}{k!^{k!}}}\right) \ \ \ \ \) <--------

The value of the second limit doesn't matter.

As I said above, my solution works... but I highly doubt that it's the intended solution.
Can anyone provide insight into what I might have missed?

.

The value of the second limit does matter. If the second limit is 0, then you will have
the \(\displaystyle \ \infty\cdot0 \ \) indeterminate form.

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

\(\displaystyle \displaystyle\lim _{k\to \infty }\left[\sqrt[k]{\bigg(\frac{e}{k!}\bigg)^{k!}} \ \right ] \ = \)

\(\displaystyle \displaystyle\lim _{k\to \infty }\left[{\bigg(\frac{e}{k!}\bigg)^{\frac{k!}{k}}} \ \right ]\ = \)

\(\displaystyle \displaystyle\lim _{k\to \infty }\left[{\bigg(\frac{e}{k!}\bigg)^{(k - 1)!}} \ \right ]\)


That limit equals 0.


So, now you have the \(\displaystyle \ \infty\cdot 0 \ \) indeterminate form to
work with in solving of the problem.

 

[Ishuda]

I've solved this problem in a very obtuse way, but I want to know if I'm missing something super obvious and there's an easier solution. I'm on quite a roll with this chapter as far as missing the obvious. The problem states:



My first step was to let \(\displaystyle a_k=\frac{\left(2k\right)!}{\left(k!\right)!}\). I then did the following:

\(\displaystyle \displaystyle ln\left(k!\right)=\sum _{x=1}^kln\left(x\right)\)

\(\displaystyle \displaystyle ln\left(k!\right)\ge \int _1^k\:ln\left(x\right)dx\)

\(\displaystyle ln\left(k!\right)\ge k\cdot ln\left(k\right)-k\)

\(\displaystyle k!\ge e^{k\cdot \:ln\left(k\right)}\cdot e^{-k}\)

\(\displaystyle k!\ge \frac{k^k}{e^k}\)

And from that, I have:

\(\displaystyle \frac{\left(2k\right)!}{\left(k!\right)!}\ge \frac{2k^{2k}}{e^{2k}}\cdot \frac{e^{k!}}{k!^{k!}}\)

So, I let \(\displaystyle b_k=\frac{2k^{2k}}{e^{2k}}\cdot \frac{e^{k!}}{k!^{k!}}\)

I have two series, such that a_k > b_k > 0 for all k >1 so I used the limit comparison test. But I first needed to prove that b_k diverges, so I used the root test on it.

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

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

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

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

The value of the second limit doesn't matter. Even if it goes to infinity, infinity times infinity is not an indeterminate form. Thus, by the root test, the series of b_k diverges. And since the series of a_k is greater than a divergent series, it too must diverge.

As I said above, my solution works... but I highly doubt that it's the intended solution. Can anyone provide insight into what I might have missed?

Why don't you just do the ratio test?

 

[ksdhart]

.

The value of the second limit does matter. If the second limit is 0, then you will have
the \(\displaystyle \ \infty\cdot0 \ \) indeterminate form.

...

That limit equals 0.

Argh. You're right. Thank you for pointing out my error. However, I still haven't the slightest clue how to actually solve this problem. This is all the guidance I have from the answer in the back of the book:

51. Each term of the series is positive, and it converges by the ratio test.

So, not only was my logic wrong about the indeterminate form, my conclusion was wrong too. The book mentions the ratio test, but I get stuck when I try that:

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

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

...and I don't know how to continue further because of the nested factorials. Intuitively, it makes sense to me that the denominator grows way faster than the numerator, and the two terms with 2k won't make a difference, so the limit goes to 0. But I can't just assume that, I have to prove it.

 

[Ishuda]

Argh. You're right. Thank you for pointing out my error. However, I still haven't the slightest clue how to actually solve this problem. This is all the guidance I have from the answer in the back of the book:



So, not only was my logic wrong about the indeterminate form, my conclusion was wrong too. The book mentions the ratio test, but I get stuck when I try that:

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

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

...and I don't know how to continue further because of the nested factorials. Intuitively, it makes sense to me that the denominator grows way faster than the numerator, and the two terms with 2k won't make a difference, so the limit goes to 0. But I can't just assume that, I have to prove it.

Lets look at an example: Suppose k is 3 then
k term: (3!)! = 6! = 1*2*3*...3!
k+1 term: (4!)! = 24! = 1*2*3*...3!*(3!+1)*(3!+2)*(3!+3)*...*22*23*4!
So what is left if we just look at ratio of the 4th to 3rd term is
\(\displaystyle \frac{7 * 8}{(3!+1)*(3!+2)*(3!+3)*...*22*23*4!}=\frac{7 * 8}{7*8*9*...*22*23*4!}=\frac{1}{9*...*22*23*24}\)

Now do the same sort of thing for the generic k.

 

[ksdhart]

Okay. That makes a lot of sense. But I can only hope that nothing this hard shows up on the test. I don't know about anybody else, but it seems like every math book I've ever had has featured one or two problems in each section that are like 1000 times harder than any other problem.