Ratio test inconclusive, what now?

[schoolslave]

Homework Statement

Determine divergence/convergence of this series:

\sum\frac{(2n)!}{(n!)^{2}} * (\frac{1}{4})^{n}

Homework Equations

ratio test?

The Attempt at a Solution

I attempted to use the ratio test and the resulting limit was 1, which means the ratio test is inconclusive. So far, I have learned to use only the ratio test with series involving factorials and powers (of n); however I have also been taught the other common dvgs/svgs tests (root, alternating, integral, etc).

I honestly have no idea on where to go from here. I tried researching online, but I only found references to sup and inf limits and I don't even know what those are.

 

[micromass]

The ratio test is conclusive with me. Could you post your calculations??

 

[Dick]

The ratio test is conclusive with me. Could you post your calculations??

It looks inconclusive to me. I get the ratio to be (2n+2)*(2n+1)/((n+1)*(n+1)*4). So ratio goes to 1. Or am I doing it wrong too?

 

[micromass]

It looks inconclusive to me. I get the ratio to be (2n+2)*(2n+1)/((n+1)*(n+1)*4). So ratio goes to 1. Or am I doing it wrong too?

Aaaargh, no, I made a mistake. It is inconclusive Thanks Dick!

 

[Dick]

Homework Statement

Determine divergence/convergence of this series:

\sum\frac{(2n)!}{(n!)^{2}} * (\frac{1}{4})^{n}

Homework Equations

ratio test?

The Attempt at a Solution

I attempted to use the ratio test and the resulting limit was 1, which means the ratio test is inconclusive. So far, I have learned to use only the ratio test with series involving factorials and powers (of n); however I have also been taught the other common dvgs/svgs tests (root, alternating, integral, etc).

I honestly have no idea on where to go from here. I tried researching online, but I only found references to sup and inf limits and I don't even know what those are.

I would pull out a version of Stirling's approximation and try to figure out what the series really looks like for large n.

 

[schoolslave]

I would pull out a version of Stirling's approximation and try to figure out what the series really looks like for large n.

What exactly is Stirling's approximation?

 

[Dick]

What exactly is Stirling's approximation?

Try this. http://en.wikipedia.org/wiki/Stirling's_approximation

 

[schoolslave]

So I would substitute the stirling approximation for every n!, correct?

 

[jbunniii]

Are you familiar with Raabe's ratio test? It works in some cases where the standard ratio test is inconclusive. If the terms of a series a_k are all positive, then compute

L = \lim_{k \rightarrow \infty} k\left(\frac{a_k}{a_{k+1}} - 1\right)

If the limit exists, then L < 1 implies convergence, and L > 1 implies divergence of the series

\sum_{k=1}^{\infty}a_k

L = 1 is inconclusive.

Raabe's test is a special case of Kummer's test. The proof is pretty easy. See, e.g., Thomson, Bruckner, and Bruckner, Elementary Real Analysis.

 

[Bohrok]

This problem intrigues me and has come to bother me since every test I've tried comes out as inconclusive (or go to 0 or ∞ and these cases aren't mentioned in the fine print for each one) using tests I learned from calc II (ratio test, root test, limit comparison test). Haven't found a good candidate for the comparison test either... Is it even possible with these tests?

 

[Dick]

So I would substitute the stirling approximation for every n!, correct?

Try it! jbunniii's suggestion of the Raabe test is a great idea too.

 

[schoolslave]

Alright, so I, hopefully correctly , applied Raabe's test and got -\frac{1}{2}, therefore the series is divergent, correct?

I also put the sum into WolframAlpha using the Stirling approximation and it just states that the series does not converge.

 

[jbunniii]

I found a more elementary solution, assuming I didn't make a mistake. Consider taking the log of each term:

\log a_n = \log\left(\frac{(2n)!}{(n!)^2} \left(\frac{1}{4}\right)^n\right)

This allows you to simplify the expression significantly. You know that convergence is impossible unless

\lim_{n \rightarrow \infty} a_n = 0

or equivalently

\lim_{n \rightarrow \infty} \log(a_n) = -\infty

If this isn't true, then you know immediately that the series diverges.

 

[jbunniii]

Alright, so I, hopefully correctly , applied Raabe's test and got -\frac{1}{2}, therefore the series is divergent, correct?

I also put the sum into WolframAlpha using the Stirling approximation and it just states that the series does not converge.

I got +1/2, not -1/2, but I might have made a mistake. If you post your work, I'll compare notes with you.

P.S. You probably already noticed, but I got the inequalities reversed in my post about Raabe's test. L > 1 implies convergence, and L < 1 implies divergence. So either +1/2 or -1/2 give the same conclusion.

 

[Dick]

I found a more elementary solution, assuming I didn't make a mistake. Consider taking the log of each term:

\log a_n = \log\left(\frac{(2n)!}{(n!)^2} \left(\frac{1}{4}\right)^n\right)

This allows you to simplify the expression significantly. You know that convergence is impossible unless

\lim_{n \rightarrow \infty} a_n = 0

or equivalently

\lim_{n \rightarrow \infty} \log(a_n) = -\infty

If this isn't true, then you know immediately that the series diverges.

If you use the Stirling approximation you can conclude the limiting behavior of the series proportional to a power of n for large n. Or you can guess this by using numerical experiments. They'll show you it does approach 0. But like what power? Answering that question will tell you whether the series converges.

 

[jbunniii]

I found a more elementary solution, assuming I didn't make a mistake.

Sure enough, I made a mistake - scratch that. The terms go to zero, as Dick said.

After a bit more scribbling, I found an elementary argument that works. It turns out to be quite simple, actually.

Write out the factors of (2n)!

The product of the even factors is 2^n n!. The product of the odd factors is

(2n-1)(2n-3)\cdots 3 \geq (2n-2)(2n-4)\cdots2 = 2^{n-1}(n-1)(n-2)\cdots 1 = 2^{n-1}(n-1)!

Thus

(2n)! \geq 2^n 2^{n-1} n! (n-1)!

from which you can easily show that

\frac{(2n)!}{(n!)^2} \frac{1}{4^n} \geq \frac{1}{2n}

and thus the series diverges by comparison with the harmonic series.

 

[Dick]

Sure enough, I made a mistake - scratch that. The terms go to zero, as Dick said.

After a bit more scribbling, I found an elementary argument that works. It turns out to be quite simple, actually.

Write out the factors of (2n)!

The product of the even factors is 2^n n!. The product of the odd factors is

(2n-1)(2n-3)\cdots 3 \geq (2n-2)(2n-4)\cdots2 = 2^{n-1}(n-1)(n-2)\cdots 1 = 2^{n-1}(n-1)!

Thus

(2n)! \geq 2^n 2^{n-1} n! (n-1)!

from which you can easily show that

\frac{(2n)!}{(n!)^2} \frac{1}{4^n} \geq \frac{1}{2n}

and thus the series diverges by comparison with the harmonic series.

Very nice. Stirling will give you the more accurate estimate that series is asymptotically c/sqrt(n). So it diverges by a limit comparison test. But I like your elementary approach.

 

[jbunniii]

Very nice. Stirling will give you the more accurate estimate that series is asymptotically c/sqrt(n). So it diverges by a limit comparison test. But I like your elementary approach.

Yes, Stirling's approximation is handy here, but I was a bit puzzled that the problem would call for it or for Raabe's test when apparently these hadn't been covered yet in the OP's studies, so I was happy to find a simple solution. Plus it's a nice trick that could come in handy again sometime, thus becoming a method instead of a trick.