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

In summary, using the Ratio Test, we can determine that the series \Sigma from n=0 to infinity (-10)n/n! is absolutely convergent and therefore convergent.
  • #1
lilypetals
12
0

Homework Statement


[tex]\Sigma[/tex] 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 [tex]\Sigma[/tex]|an| is convergent.

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

The Ratio Test, which states that if the limit as n goes to infinity of |an+1/an| 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 an 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:

[tex]\Sigma[/tex] n=0 to infinity |(-10)n|/n!

So, I need to consider a bn which is greater than an, which converges.

This is where I get stuck. I haven't been able to find a bn for which I could solve the limit as n goes to infinity. Anyone have a good method for this?
 
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  • #2
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?
 
  • #3
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?
 
  • #4
Use the ratio test. an = 10^n/(n!). What is an+1/an.
 
  • #5
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.
 
  • #6
If I use the ratio test...|an+1/an|=|((-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?
 
  • #7
lilypetals said:
If I use the ratio test...|an+1/an|=|((-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.
 
  • #8
Thank you!
 

1. What does it mean for a series to converge or diverge?

When a series converges, it means that the sum of all the terms in the series approaches a finite value as the number of terms increases. On the other hand, a series diverges if the sum of the terms either approaches infinity or does not have a finite limit.

2. What is the difference between absolute convergence and convergence?

Absolute convergence refers to a series in which the absolute values of the terms converge, while convergence refers to a series where the terms themselves converge. A series can be absolutely convergent without being convergent, but a convergent series is always absolutely convergent.

3. How do you determine if a series is absolutely convergent?

To determine if a series is absolutely convergent, you can use the ratio or root test. If the limit of the absolute values of the terms is less than 1, the series is absolutely convergent. If the limit is greater than 1, the series is divergent. If the limit is equal to 1, the test is inconclusive and another method must be used.

4. What is the significance of testing for absolute convergence?

Testing for absolute convergence is important because it guarantees that the series will converge regardless of the order in which the terms are added. This is useful when rearranging terms in a series, as the sum will remain the same if the series is absolutely convergent.

5. Can a series be both absolutely convergent and conditionally convergent?

Yes, a series can be both absolutely convergent and conditionally convergent. This means that the series is absolutely convergent, but when the absolute values of the terms are not considered, the series still converges. An example of this is the alternating harmonic series.

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