Convergence of a Complex Series

[Draconifors]

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]

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?

 

[wabbit]

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.

 

[HallsofIvy]

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!