# I am unsure of factorials

1. Jun 20, 2010

### PhysicsUnderg

I am unsure as to how factorials should be expanded.

I have $$\sum$$$$\stackrel{1}{(2n!)}$$ (if what was just typed did not make sense due to html error on my part, it is supposed to say the sum of 1/(2n)!) from n=1 to infinity. I did the ratio test and found the limit to be 0, which is less than 1, which means this series is absolutely convergent. My fear is that I didn't expand the factorials properly and perhaps my answer is wrong. Could someone let me know if my answer makes sense?

FYI: (2n+2)! I expanded as (2n+2)(2n+1)(2n)!

2. Jun 20, 2010

### djeikyb

You can always try another method.
For example, the series of 1/(2n)! goes
1/2, 1/24, 1/720, 1/40320.

Compare it to the series 1/2^n.
1/2, 1/4, 1/8, 1/16, 1/32, and so on

Now, notice that every element of the second series is greater than or equal to the corresponding element of your series.
1/2 = 1/2
1/4 > 1/24
1/8 > 1/720

Now, 1/2^n is a convergent series, yes?

So if you are summing these series, you can compare the two using what we always called racetrack theorem. Every term of my series is greater than the corresponding term of your series. My series converges to 1 as n->infinity. Can your series possibly be larger?

3. Jun 20, 2010

### PhysicsUnderg

Well, I know from the comparison test, if An=1/(2n)! and Bn=1/2n and if An is less than Bn, and if Bn is convergent, than An is also convergent. Yet, in this case, Bn is divergent (because 1/2n is a p-series with p=1, and this is divergent.) So, the comparison test shows nothing...

So, I am guess I still am not seeing the answer. :-(

4. Jun 20, 2010

### Staff: Mentor

You did it just right.

Here's a tip on the LaTeX - don't use \stackrel for fractions - use \frac{}{}. Click on what I've written below to see the LaTeX code I wrote.
$$\sum_{n = 1}^{ \infty} \frac{1}{(2n)!}$$

5. Jun 20, 2010

### djeikyb

The comparison you are trying An=1/(2n)! and Bn=1/2n is not showing anything because Bn is divergent. Instead of these, try it with a different comparison. The one I mentioned before works.
An=1/(2n)! and Cn=1/2^n
An < Cn and Cn is convergent, so An must also be convergent.

6. Jun 21, 2010

### PhysicsUnderg

To djeikyb: your response makes sense now. I misunderstood 1/2^n to be 1/2n. Thank you. :-)

To Mark44: Thank you :-)

7. Jun 21, 2010

### HallsofIvy

Staff Emeritus
Notice that (2n!), which is what you originally wrote, and (2n)! are very different!

$$\sum \frac{1}{2n!}= \frac{1}{2}\sum \frac{1}{n!}= \frac{1}{2}e$$.