Proving That The Series is Convergent or Divergent

jsewell94
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Homework Statement



Determine whether the following series converges or diverges:
\sum_{}^{} ( \frac{1}{3} )^{ln(n)}

Homework Equations



N/A

The Attempt at a Solution



See attached document..

I had my Calc 2 final today, and this was our hard problem...but I don't know if my method is valid or not. Could you help me determine if it is? And if it's not, could you tell me where the flaw in my reasoning is/a better method?
 

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jsewell94 said:

Homework Statement



Determine whether the following series converges or diverges:
\sum_{}^{} ( \frac{1}{3} )^{ln(n)}

Homework Equations



N/A

The Attempt at a Solution



See attached document..

I had my Calc 2 final today, and this was our hard problem...but I don't know if my method is valid or not. Could you help me determine if it is? And if it's not, could you tell me where the flaw in my reasoning is/a better method?

It looks ok to me. You could shorten the whole argument up a lot. 3^{ln(n)}=(e^{ln(3)})^{ln(n)}=e^{ln(3) ln(n)}=(e^{ln(n)})^{ln(3)}=n^{ln(3)}. Just use ln(3) for k.
 
Dick said:
It looks ok to me. You could shorten the whole argument up a lot. 3^{ln(n)}=(e^{ln(3)})^{ln(n)}=e^{ln(3) ln(n)}=(e^{ln(n)})^{ln(3)}=n^{ln(3)}. Just use ln(3) for k.

Oh wow, that makes sense!

Yeah, I just..couldn't figure out the easy way in the 15 minutes I had left on the test :(
Yay! That means I was the only one who got it :D
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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