Power Series with a funky denominator

[NastyAccident]

Homework Statement

Find the radius of convergence and interval of convergence of the following series.
$$\sum^{\infty}_{n=1}\frac{n!*x^{n}}_{5*11*17*\cdots*(6n-1)}$$

Homework Equations

Knowledge of Power Series.
Factorials?

The Attempt at a Solution

I'm really unsure where to start, however, I did attempt [see the attached pdf] solving the series by just attacking it head on with the ratio test. After I obtained the convergence interval, I reran the ratio test using x=6 and received an inconclusive result (L=1).

The only type of Power Series that I've dealt with similarly was:

$$\sum^{\infty}_{n=1}\frac{n^{2}x^{n}}_{2*4*6*\cdots*(2n)}$$

$$2*4*6*\cdots*(2n) = (2*1)*(2*2)*(2*3)*\cdots*(2*n) = 2^{n}(1*2*3*\cdots*n)=2^n*n!$$

However, with this particular series, as I said earlier, I'm not really sure how to handle the series because of the type of denominator I'm dealing with... Heck, I'm not even sure what this type of denominator is called.

ANY, and ALL help will be appreciated and thanked!



NastyAccident.

 

[Mark44]

Try the ratio test. That will probably show that this series converges, and the ratio can be used to find the values of x for which it converges. That gives you your interval of convergence and the radius of convergence. You'll probably need to check the endpoints of the interval separately.

 

[NastyAccident]

Try the ratio test. That will probably show that this series converges, and the ratio can be used to find the values of x for which it converges. That gives you your interval of convergence and the radius of convergence. You'll probably need to check the endpoints of the interval separately.

So far I've done all of that. (Please see the attached PDF. Sorry that I couldn't save it as a GIF/JPG since I currently am using MathType-Lite for Mac so I had to export my pages document containing the MathType-lite equation to a PDF.)

*Shown in PDF on original post*
R = 6
I = -6<x<6

I am having trouble with actually checking the endpoints determined since I'm unsure what the denominator actually is.

For instance, I know that if x = -6 it is an alternating series.
Thus, I should run the Alternating Series Test (AST). However, since I'm unsure what the actual denominator is. This causes me to not be able to justify the limit (part II of the AST). Justifying a_n+1 < a_n (Part I of the AST) is easy assuming that I times the denominator by (6n+1) so it becomes

5*11*17*...*(6n+1)*(6n-1).

With x = 6, I'm not even sure what test to begin with. Since I can't justify the limit... [Normally, I run the Test of Divergence first.]
NastyAccident

 

[Mark44]

Your attachment is still pending approval, so my previous post applied to what I could see in your post.

For x = -6, the general term in your series is
$$\frac{n! (-6)^n}{5 * 11 * 17 * ... * (6n - 1)}$$

This can be rewritten as
$$\frac{6*1}{5}~\frac{6*2}{11}~\frac{6*3}{17}...~\frac{6n - 6}{6n - 7}~\frac{6n}{6n - 1}$$

I have omitted the (-1)^n factor, since we already know this is an alternating series, and the goals are to determine whether the general term a_n in this series approaches 0 and a_{n + 1} < a_n.

Pretty clearly the series is NOT monotone decreasing. To get a_{n + 1}, we multiply a_n by 6(n + 1)/(6n + 5), a number that is larger than 1. From this, we see that the sequence of partial sums does not approach zero, so the series diverges.

The same is true when x = 6: the series diverges at this endpoint of the convergence interval.

 

[NastyAccident]

Your attachment is still pending approval, so my previous post applied to what I could see in your post.

For x = -6, the general term in your series is
$$\frac{n! (-6)^n}{5 * 11 * 17 * ... * (6n - 1)}$$

This can be rewritten as
$$\frac{6*1}{5}~\frac{6*2}{11}~\frac{6*3}{17}...~\frac{6n - 6}{6n - 7}~\frac{6n}{6n - 1}$$

I have omitted the (-1)^n factor, since we already know this is an alternating series, and the goals are to determine whether the general term a_n in this series approaches 0 and a_{n + 1} < a_n.

Pretty clearly the series is NOT monotone decreasing. To get a^{n + 1}, we multiply a_n by 6(n + 1)/(6n + 5), a number that is larger than 1. From this, we see that the sequence of partial sums does not approach zero, so the series diverges.

The same is true when x = 6: the series diverges at this endpoint of the convergence interval.

Ah, my apologies in regards to uploading a PDF I wasn't aware of the approval requirement. Thank you for help so far as well!

Now, you converted the numerator to a form similar to the denominator and then evaluated it.

$$\frac{6*1}{5}~\frac{6*2}{11}~\frac{6*3}{17}...~\frac{6n - 6}{6n - 7}~\frac{6n}{6n - 1}$$

$$\frac{6}{5}~\frac{12}{11}~\frac{18}{17}...~\frac{6n - 6}{6n - 7}~\frac{6n}{6n - 1}$$

This makes sense so far in my mind.

To get a_n+1 you took the last term $$\frac{6n}{6n - 1}$$ and plugged in for n = n+1. So, a_n+1=$$...*\frac{6(n+1)}{6(n+1) - 1}=...*\frac{6n+6)}{6n+6) - 1}=...*\frac{6n+6)}{6n+5}$$

Then you were able to see that the limit as n->infinity = 1 (not zero), therefore it is not convergent by the Test of Divergence for both x=-6,6. Correct?



NastyAccident

 

[Mark44]

Yes. What you have for a_n is the product of n factors, each of which is slightly larger than 1, hence lim a_n is not zero.

 

[NastyAccident]

Yes. What you have for a_n is the product of n factors, each of which is slightly larger than 1, hence lim a_n is not zero.

Thanks! I appreciate it =)



NastyAccident