# Testing convergence of Series

## Homework Statement

"Determine whether the following series converge or diverge. If the series is geometric or telescoping, find its sum.":

## \left ( \sum_{k=1}^\infty2^{3k} *3^{1-2k} \right)##

## Homework Equations

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The different tests for convergence?

## The Attempt at a Solution

Ok, I've looked at all the tests for convergence I know, and as far as I've been able to tell, none of them can work with this? Here's a quick outline about each of the tests and why I don't think I can use it.

Divergence Test: I have to say I can't even do the limit of this. It just gives me ##2^{infinity}*3^{1-infinity}##, no?
Geometric series: I don't have only one exponent, so I can't do this. Neither of the exponents looks like ##(n-1)## so I don't know.
Telescoping series: There's nothing that cancels out. I also can't take the limit.
p-Series: It's not in the form ##{\frac {1} {p}}##.
Alternating series: It doesn't have a ##-1^n## term.
Integral test: ##\int 2^{3k} *3^{1-2k} \, dx## . I can't integrate this because of the ##k## being a power.
Root test: While both are raised to a power of ##k##, I don't think I can apply the root test because of the ##1-2k##.
Ratio test: This is usually my go to, and I thought it would work. However, I get ##\frac {1} {9}## and then the ##3^{1-2k}## just stays there.
Comparison test: Even if I wanted to use the limit or direct comparison, I wouldn't even know what to compare it with.

The solutions tell me it's supposed to converge to 24 and to use the geometric test. However, I really don't see how the geometric test is supposed to apply.

Thanks for your time, and I hope I did everything right in posting this!

wabbit
Gold Member
This is simpler than what you're trying. Go back to the expression under the sum and try to see if you can put it in a simpler form first. Use the relations you know about exponents and multiplication, such as ## x^{nk}=(x^n)^k ##, or ## x^{p+q}=x^px^q ##, etc.

Draconifors
This is simpler than what you're trying. Go back to the expression under the sum and try to see if you can put it in a simpler form first. Use the relations you know about exponents and multiplication, such as ## x^{nk}=(x^n)^k ##, or ## x^{p+q}=x^px^q ##, etc.

Ooh, that helped!
Ok, so doing that I get ##(2^3)^k*3^1*3^{-2k}##. That becomes ##3*8^k*3^{-2k}##. I can see where the 24 is supposed to come from: 3*8. I just can't get it to simplify to that.
I rearrange the expression so it's ##\frac{3*8^k}{3^{2k}}## and then that becomes ##({\frac{8}{9}})^k *3## If the summation started at k=0, I would see that this is a geometric sum with a = 3, r = 8/9. However, the summation starts at k=1, so shouldn't I have ##n-1## as an exponent?

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wabbit
Gold Member
Yep, you got it - about the starting point ## k=1 ## , you just need to be careful when determining the actual sum, using things like ## \sum_{k\geq 1}a_k=(\sum_{k\geq 0}a_k)-a_0 ##, or in your case using ## a^k=a\cdot a^{k-1} ## : while your series may not be exactly the standard geometric sum, you can express in terms of that.

Draconifors
HallsofIvy
Homework Helper
Ooh, that helped!
Ok, so doing that I get ##(2^3)^k*3^1*3^{-2k}##. That becomes ##3*8^k*3^{-2k}##. I can see where the 24 is supposed to come from: 3*8. I just can't get it to simplify to that.
I rearrange the expression so it's ##\frac{3*8^k}{3^{2k}}## and then that becomes ##({\frac{8}{9}})^k *3## If the summation started at k=0, I would see that this is a geometric sum with a = 3, r = 8/9. However, the summation starts at k=1, so shouldn't I have ##n-1## as an exponent?
Either change the exponent or subtract off the "k= 0" term.

Draconifors
Alright, thank you very much to the two of you, this really helped me out!

Have a great day!