Determining the absolute convergence, convergence, or divergence of a series.

[lilypetals]

Homework Statement

\Sigma from n=0 to infinity (-10)ⁿ/n!
Determine the absolute convergence, convergence, or divergence of the series.

Homework Equations

In this section, it's suggested that we use the following to determine a solution:

A series is called absolutely convergent if the series of absolute values \Sigma|a_n| is convergent.

The Comparison Test, which states that if the series b_n is convergent and greater than the series a_n, then the series a_n is also convergent; and if the series b_n is divergent and less than the series a_n, then the series a_n is also divergent.

The Ratio Test, which states that if the limit as n goes to infinity of |a_n+1/a_n| is less than 1 the series is absolutely convergent; if it is greater than 1 the series is divergent; and if it is equal to 1 the Ratio Test is inconclusive.

The Root Test, which states that if the limit as n goes to infinity of the nth root of a_n is less than 1 the series is absolutely convergent; if it is greater than 1 the series is divergent; and if it is equal to 1 the Root Test is inconclusive.

The Attempt at a Solution

I decided that the simplest method would be to apply the Comparison Test to the series of absolute values:

\Sigma n=0 to infinity |(-10)ⁿ|/n!

So, I need to consider a b_n which is greater than a_n, which converges.

This is where I get stuck. I haven't been able to find a b_n for which I could solve the limit as n goes to infinity. Anyone have a good method for this?

 

[vela]

Usually when you have n appears in factorials and in the exponent, the ratio test is a good one to try. That said, you can prove the series converges using the comparison test. What series have you considered and where do you get stuck with them?

 

[lilypetals]

I guess that's the problem--I'm not even sure which series to consider. I know that if I guess that the given series converges, I need a larger series which also converges, and I can meet the criteria by making the denominator smaller or the numerator larger. It's hard to imagine making the numerator larger, so I considered the series |(-10)ⁿ|/n, and tried taking the limit as n goes to infinity, but that limit is infinite, isn't it?

 

[deluks917]

Use the ratio test. a_n = 10^n/(n!). What is a_n+1/a_n.

 

[vela]

Yeah, you need a series that converges. Keep in mind you don't have to find a series which is greater than the original series for all n; it just has to be greater for n>N where N is some finite number. That's because you can always discard a finite number of terms when testing for convergence. You only care what happens as n goes to infinity.

In this problem, you can see the numerator always increases by a factor of 10, but the bottom grows at a faster rate. What if you compare it to something like k(1/10)ⁿ where the top and bottom always change by the same factor? You can set the constant k to whatever you want, since it won't affect the second series' convergence, to make the comparison easier to prove.

 

[lilypetals]

If I use the ratio test...|a_n+1/a_n|=|((-10)ⁿ⁺¹/(n+1)!)/(-10)ⁿ/n!|=10/(n+1).

So the limit as n goes to infinity of 10/(n+1) can be found by dividing by n, and taking the limit of the result.

And the limit of n as it goes to infinity of (10/n)/(1+1/n)=0/(1+0)=0.

Since this is less than one, the Ratio Test says that the series (-10)ⁿ/n! is absolutely convergent, and therefore convergent.

Is this correct?

 

[Dick]

If I use the ratio test...|a_n+1/a_n|=|((-10)ⁿ⁺¹/(n+1)!)/(-10)ⁿ/n!|=10/(n+1).

So the limit as n goes to infinity of 10/(n+1) can be found by dividing by n, and taking the limit of the result.

And the limit of n as it goes to infinity of (10/n)/(1+1/n)=0/(1+0)=0.

Since this is less than one, the Ratio Test says that the series (-10)ⁿ/n! is absolutely convergent, and therefore convergent.

Is this correct?

Yes, that's it.

 

[lilypetals]

Thank you!