Did I correctly prove the divergence of this series?

[mathnoobie]

Homework Statement

Verify that the infinite series diverges.
I have the series from n=1 to infinity of (2^(n)+1/2^(n+1)

Homework Equations

Nth term test(This is the way the book did it but I did it used the geometric series test
and I just want to verify if my Algebra was correct)

The Attempt at a Solution

First I split the series into two separate series and let the series go from n=0 to infinity.
So I have 2^(n+1)/2^(n+2) add 1/2^(n+2).
I believe I can then bring down the some of the exponents and simplify to
2*2^n/4*2^n add 1/4*2^n
so for the first series I let a=1/2 and r=1^n
because |r|=|1| is greater than or equal to 1, the series diverges.
I completely ignored the second series that I made because its irrelevant to simplify
as I already know the first series diverges.

Is this proof valid and was my algebra correct?

 

[Mark44]

Homework Statement

Verify that the infinite series diverges.
I have the series from n=1 to infinity of (2^(n)+1/2^(n+1)

Is this your series?

$$ \sum_{n = 1}^{\infty} \left(2^n + \frac{1}{2^{n+1}}\right)$$

You are missing a right paren, so I'm not sure what you intended.

The easiest approach is the n-th term test for divergence. It seems like a lot of extra work to try to make this series look like a geometric series, for which the general term is arⁿ.

Homework Equations

Nth term test(This is the way the book did it but I did it used the geometric series test
and I just want to verify if my Algebra was correct)

The Attempt at a Solution

First I split the series into two separate series and let the series go from n=0 to infinity.
So I have 2^(n+1)/2^(n+2) add 1/2^(n+2).
I believe I can then bring down the some of the exponents and simplify to
2*2^n/4*2^n add 1/4*2^n
so for the first series I let a=1/2 and r=1^n
because |r|=|1| is greater than or equal to 1, the series diverges.
I completely ignored the second series that I made because its irrelevant to simplify
as I already know the first series diverges.

Is this proof valid and was my algebra correct?

 

[mathnoobie]

Is this your series?

$$ \sum_{n = 1}^{\infty} \left(2^n + \frac{1}{2^{n+1}}\right)$$

You are missing a right paren, so I'm not sure what you intended.

The easiest approach is the n-th term test for divergence. It seems like a lot of extra work to try to make this series look like a geometric series, for which the general term is arⁿ.

(2^(n)+1)/2^(n+1)

I apologize for the confusion, that is my series.
I agree that the easiest approach is the n-th term test, however I don't always see the easiest approach every single time.
The problem with series is that you can do the wrong thing and still get the right answer, I just want to make sure my proof was right
since I got the right answer.

 

[micromass]

I agree that the easiest approach is the n-th term test, however I don't always see the easiest approach every single time.

I think that if you're given a series, that the first thing you should always try is the n-th term test. If you don't find it using that, then you can move on. But the n-th term test is usually a lot easier than the other tests.

 

[mathnoobie]

I think that if you're given a series, that the first thing you should always try is the n-th term test. If you don't find it using that, then you can move on. But the n-th term test is usually a lot easier than the other tests.

Thank you for the advice, I will always do that from now on. I kinda saw that I should've done that halfway through the problem but I thought to myself why not finish it.

Could anyone though please confirm if my Algebra was correct? I have a test coming up and there will be geometric series that require this kind of algebraic manipulation and I just want to know if everything I did algebraically was correct.
I mostly just need to know if the exponent rule I did was right.
Like when you have 2^(n+2), its actually 2^2 * 2^n. This is a valid right?

 

[Mark44]

(2^(n)+1)/2^(n+1)

Like so:
So this is your series.
$$ \sum_{n = 1}^{\infty} \left( \frac{2^n + 1}{2^{n+1}} \right)$$

I apologize for the confusion, that is my series.
I agree that the easiest approach is the n-th term test, however I don't always see the easiest approach every single time.

But you should try the easy approaches first, and the n-th term test for divergence is probably the easiest of all the tests.

The problem with series is that you can do the wrong thing and still get the right answer, I just want to make sure my proof was right
since I got the right answer.

Here's from your first post...

First I split the series into two separate series and let the series go from n=0 to infinity.
So I have 2^(n+1)/2^(n+2) add 1/2^(n+2).

Wouldn't it have been simpler to leave the index starting from 1?
Then you would have 2^n/(2^(n + 1)) + 1/(2^(n + 1)).

I believe I can then bring down the some of the exponents and simplify to
2*2^n/4*2^n add 1/4*2^n

Use + instead of "add".
2^n/(2^(n + 1)) + 1/(2^(n + 1)) = 1/2 + 1/(2^(n + 1)).

The first part of what you have is correct, but can be greatly simplified.
2*2^n/4*2^n = 1/2

so for the first series I let a=1/2 and r=1^n

No. For a geometric series r has to be a constant. Of course, for any finite n, 1^n is 1, so was this a typo?

Your first series is $$ \sum_{n = 1}^{\infty} \frac{1}{2} $$
This is not a geometric series.

because |r|=|1| is greater than or equal to 1, the series diverges.
I completely ignored the second series that I made because its irrelevant to simplify
as I already know the first series diverges.

 

[micromass]

Thank you for the advice, I will always do that from now on. I kinda saw that I should've done that halfway through the problem but I thought to myself why not finish it.

Could anyone though please confirm if my Algebra was correct? I have a test coming up and there will be geometric series that require this kind of algebraic manipulation and I just want to know if everything I did algebraically was correct.
I mostly just need to know if the exponent rule I did was right.
Like when you have 2^(n+2), its actually 2^2 * 2^n. This is a valid right?

Your algebra was perfectly fine.

But there was a mistake in the proof: you split the series in two parts

$$\sum{a_n+b_n}=\sum a_n +\sum b_n$$

and deduce from that that a series in the right hand side diverges, thus the left-hand side diverges as well. This is not true. For example

$$\sum{1-1}=\sum 1 - \sum 1$$

the two series in the right-hand side diverge, but the left-hand side does not diverge.

The rule

$$\sum{a_n+b_n}=\sum a_n +\sum b_n$$

only holds if both $\sum a_n$ as $\sum b_n$ converge. If both diverge, then there is nothing that can be said.

 

[mathnoobie]

Thank you guys so much for the thorough explanations! I really need to review my property of series. My biggest concern was the algebra but thankfully I did that right.
As for the use of add, I thought it would be easier to read if I used add instead of +(not sure why).

 

[HallsofIvy]

Your algebra was perfectly fine.

But there was a mistake in the proof: you split the series in two parts

$$\sum{a_n+b_n}=\sum a_n +\sum b_n$$

and deduce from that that a series in the right hand side diverges, thus the left-hand side diverges as well. This is not true. For example

$$\sum{1-1}=\sum 1 - \sum 1$$

the two series in the right-hand side diverge, but the left-hand side does not diverge.

The rule

$$\sum{a_n+b_n}=\sum a_n +\sum b_n$$

only holds if both $\sum a_n$ as $\sum b_n$ converge. If both diverge, then there is nothing that can be said.

Well, if $a_n> 0$, $b_n> 0$, and that is the situation here, and either $\sum a_n$, or $\sum b_n$, or both diverge, then $\sum (a_n+ b_n)$ diverges.

 

[Mark44]

One point that micromass made was very pertinent - about splitting series into parts and when you can do that. That thought came to mind while I was writing a response, but I didn't include it, so I'm glad he did.