Convergence of a series with n-th term defined piecewise

In summary, the conversation discusses testing a series for convergence or divergence. The attempt at a solution involves finding the 'n'th term of the series and considering two cases. The person then checks their formulas for 'r_n' and tries grouping the odd and even terms together to determine convergence. The conversation ends with a discussion on proving absolute convergence.
  • #1
danielbaker453
26
2

Homework Statement



Test the series for convergence or divergence
##1/2^2-1/3^2+1/2^3-1/3^3+1/2^4-1/3^4+...##

Homework Equations


rn=abs(an+1/an)

The Attempt at a Solution


With some effort I was able to figure out the 'n' th tern of the series
an =
\begin{cases}
2^{-(0.5n+1.5)} & \text{if } n is odd \\
-3^{-(0.5n+1)} & \text{if } n is even
\end{cases}

Next, I considered two cases,
Case I. When an is odd,
This gives rn={(2/3)^(0.5n+1.5)}<1 for all n
Thus the terms of the series seem to decrease.
As n----->infinity, rn----->0
The series appears to be convergent.

Case II. When an is even
rn=0.5{(3/2)^(0.5n+1)}>1
As n------>infinity, rn----->infinity
This time around the series seems to be divergent.

What am I doing wrong? What should I do to properly determine the convergence/divergence of the series?
 
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  • #2
Just check your formulas for rn again. Every even term has smaller magnitude than the odd term just before it. So your statement that the even partial sums go to infinity is wrong. Since you don't really show your steps in getting those formulas, I can't say much more.
 
  • #3
FactChecker said:
Just check your formulas for rn again. Every even term has smaller magnitude than the odd term just before it. So your statement that the even partial sums go to infinity is wrong. Since you don't really show your steps in getting those formulas, I can't say much more.
These are my steps:
I considered the even term as an=-1/(3^(0.5n+1)) which makes the odd term after it an+1=1/(2^(0.5n+2).
The ratio rn=0.75*((1.5)^n) approaches infinity as n approaches infinity. This means that the series is divergent in this case.
I plugged in n+1 for n to determine the odd term following the even term.
But in the previous case, the series turned out to be convergent.
 
  • #4
FactChecker said:
Just check your formulas for rn again. Every even term has smaller magnitude than the odd term just before it. So your statement that the even partial sums go to infinity is wrong. Since you don't really show your steps in getting those formulas, I can't say much more.
Untitled.png

Here is a plot of the magnitudes of a few terms. Although what you said is true( every even term is smaller than the preceeding odd term), every odd term after an even term is larger. How can I tell now whether the series is convergent or divergent?
 

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  • #5
danielbaker453 said:
View attachment 221085
Here is a plot of the magnitudes of a few terms. Although what you said is true( every even term is smaller than the preceeding odd term), every odd term after an even term is larger. How can I tell now whether the series is convergent or divergent?

Try a comparison test. Can you think of a convergent series to compare it to? Group the odd-even terms in pairs.
 
  • #6
Dick said:
Try a comparison test. Can you think of a convergent series to compare it to? Group the odd-even terms in pairs.
If I group the odd and even terms,I obtain ##\sum_{n=2}^\infty {(1/2^n)-(1/3^n)}##
This is less than the geometric series ##\sum_{n=2}^\infty {(1/2^n)}## which is convergent. This makes my series convergent as well.
However, if the original series were conditionally convergent, wouldn't grouping change the convergence as by grouping terms in different manners we could change the convergence of the series?
Also, why does the graph show that every even term has a greater(in magnitude) odd term succeeding it(in case the series is convergent)?
 
  • #7
danielbaker453 said:
If I group the odd and even terms,I obtain ##\sum_{n=2}^\infty {(1/2^n)-(1/3^n)}##
This is less than the geometric series ##\sum_{n=2}^\infty {(1/2^n)}## which is convergent. This makes my series convergent as well.
However, if the original series were conditionally convergent, wouldn't grouping change the convergence as by grouping terms in different manners we could change the convergence of the series?
Also, why does the graph show that every even term has a greater(in magnitude) odd term succeeding it(in case the series is convergent)?

Ok, so maybe best to note the series is absolutely convergent first. There are equally easy comparisons to do that. Once you know that you can rearrange to your hearts content.
 
  • #8
Dick said:
Ok, so maybe best to note the series is absolutely convergent first. There are equally easy comparisons to do that. Once you know that you can rearrange to your hearts content.
Okay...
Now if I want to prove absolute convergence, I want to prove that the series ##\frac 1 {2^2}+\frac 1 {3^2}+\frac 1 { 2^3}+\frac 1 { 3^3}+...## is convergent.
This can be viewed as the sum of two convergent geometric series and so it will be convergent as well proving that my series absolutely convergent and therefore convergent. Am I right?
 
  • #9
danielbaker453 said:
Okay...
Now if I want to prove absolute convergence, I want to prove that the series ##\frac 1 {2^2}+\frac 1 {3^2}+\frac 1 { 2^3}+\frac 1 { 3^3}+...## is convergent.
This can be viewed as the sum of two convergent geometric series and so it will be convergent as well proving that my series absolutely convergent and therefore convergent. Am I right?

Sounds good to me.
 
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  • #10
Dick said:
Sounds good to me.
But doesn't this proof of absolute convergence also assume that rearranging the terms has no effect of the convergence?
 
  • #11
danielbaker453 said:
But doesn't this proof of absolute convergence also assume that rearranging the terms has no effect of the convergence?

Why? You didn't rearrange the series for the comparison. You just took the absolute value of all of the terms.
 
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  • #12
Dick said:
Why? You didn't rearrange the series for the comparison. You just took the absolute value of all of the terms.
I did. But when I treated this absolute value series as a sum of two series, doesn't the order in which the terms of the two series are added matter?
 
  • #13
danielbaker453 said:
I did. But when I treated this absolute value series as a sum of two series, doesn't the order in which the terms of the two series are added matter?
I mean two series could be added in many different ways. For example, first term of one with the second of the other instead of adding term by term.
 
  • #14
danielbaker453 said:
I did. But when I treated this absolute value series as a sum of two series, doesn't the order in which the terms of the two series are added matter?

It's pretty clear that the sum of two absolutely convergent series is absolutely convergent, isn't it? At which point there is no problem with order.
 
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  • #15
Dick said:
It's pretty clear that the sum of two absolutely convergent series is absolutely convergent, isn't it? At which point there is no problem with order.
Oh! I get it now. Thanks a lot!
 
  • #16
The comparison of even and odd can tell you quickly that the sum of evens is convergent, but you should also review your calculation of rn for case 2 in your original post. It must be wrong.
 
  • #17
danielbaker453 said:

Homework Statement



Test the series for convergence or divergence
##1/2^2-1/3^2+1/2^3-1/3^3+1/2^4-1/3^4+...##

Homework Equations


rn=abs(an+1/an)

The Attempt at a Solution


With some effort I was able to figure out the 'n' th tern of the series
an =
\begin{cases}
2^{-(0.5n+1.5)} & \text{if } n is odd \\
-3^{-(0.5n+1)} & \text{if } n is even
\end{cases}

Next, I considered two cases,
Case I. When an is odd,
This gives rn={(2/3)^(0.5n+1.5)}<1 for all n
Thus the terms of the series seem to decrease.
As n----->infinity, rn----->0
The series appears to be convergent.

Case II. When an is even
rn=0.5{(3/2)^(0.5n+1)}>1
As n------>infinity, rn----->infinity
This time around the series seems to be divergent.

What am I doing wrong? What should I do to properly determine the convergence/divergence of the series?

You are making the problem waaaay too hard. Just look at partial sums: if ##S_n## is the sum of the first ##n## terms, you can certainly write
$$S_{2N} = \left( 1/2^2 + 1/2^3 + \cdots + 1/2^{N+1} \right)
- \left( 1/3^2 + 1/3^3 + \cdots + 1/3^{N+1} \right)$$
There is absolutely nothing wrong here: we are not re-arranging the terms of an infinite series, we are just re-arranging the terms of a finite summation.

It is easy to examine whether or not ##S_{2N}## has a finite limit as ##N \to \infty## Furthermore, ##S_{2N+1}## differs from ##S_{2N}## by a single term that goes to zero for large ##N##, so if the even sums converge to some limit ##L## then so do the odd sums.
 
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  • #18
Ray Vickson said:
You are making the problem waaaay too hard. Just look at partial sums: if ##S_n## is the sum of the first ##n## terms, you can certainly write
$$S_{2N} = \left( 1/2^2 + 1/2^3 + \cdots + 1/2^{N+1} \right)
- \left( 1/3^2 + 1/3^3 + \cdots + 1/3^{N+1} \right)$$
There is absolutely nothing wrong here: we are not re-arranging the terms of an infinite series, we are just re-arranging the terms of a finite summation.

It is easy to examine whether or not ##S_{2N}## has a finite limit as ##N \to \infty## Furthermore, ##S_{2N+1}## differs from ##S_{2N}## by a single term that goes to zero for large ##N##, so if the even sums converge to some limit ##L## then so do the odd sums.
That's amazing! I was so focused on thinking of a test that would work that I didn't even pay attention to the very basics! Thanks for giving me this perspective.
 
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  • #19
danielbaker453 said:
That's amazing! I was so focused on thinking of a test that would work that I didn't even pay attention to the very basics! Thanks for giving me this perspective.

Well, a comparison test would certainly work as well. You could prove absolute convergence by noting that in your absolute series
$$(1/2^2+1/3^2) + (1/2^3+1/3^3) + \cdots $$ we have
$$1/2^2+1/3^2 < 2/2^2, \; 1/2^3 + 1/3^3 < 2/2^3, \ldots$$
so the series is bounded above by ##\sum 2/2^n##.
 
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1. What is the definition of convergence for a series with n-th term defined piecewise?

The definition of convergence for a series with n-th term defined piecewise is when the limit of the series approaches a finite value as n approaches infinity. This means that as more terms are added to the series, the sum becomes closer and closer to a specific number.

2. How do I determine if a series with n-th term defined piecewise converges or diverges?

To determine if a series with n-th term defined piecewise converges or diverges, you can use the comparison test, ratio test, or the integral test. These methods involve comparing the series to known convergent or divergent series or integrating the series to determine its convergence.

3. Can a series with n-th term defined piecewise converge and diverge at the same time?

No, a series with n-th term defined piecewise can only either converge or diverge. It cannot do both at the same time. This is because the definition of convergence is when the limit of the series approaches a specific value, while divergence is when the limit does not exist or approaches infinity.

4. Does the order of the terms in a series with n-th term defined piecewise affect its convergence?

Yes, the order of the terms in a series with n-th term defined piecewise can affect its convergence. If the terms are rearranged, the resulting sum may be different and can lead to a different convergence or divergence outcome. However, if the series is absolutely convergent, the order of the terms does not matter.

5. Are there any special cases for the convergence of a series with n-th term defined piecewise?

Yes, there are some special cases for the convergence of a series with n-th term defined piecewise. For example, if the series has alternating signs and the terms decrease in magnitude, the series will converge. Additionally, if the series has a telescoping sum, where most terms cancel each other out, it will also converge. However, these are not the only special cases and it is important to use proper convergence tests to determine convergence or divergence.

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