Series convergence, root test

  • #1

Homework Statement



Does it converge, and what is the sum:

[tex] \sum_{n=1}^{\infty}\frac{1}{n n^{\frac{1}{n}}} [/tex]

Homework Equations





The Attempt at a Solution



Please check my method and conclusion:

Using the root test:

[tex] \displaystyle\lim_{n\to\infty}\left|\frac{1}{n n^{\frac{1}{n}}} \right|^{\frac{1}{n}} = \displaystyle\lim_{n\to\infty}\frac{1}{n n^{\frac{1}{n}}} = \displaystyle\lim_{n\to\infty}\frac{1}{n}\displaystyle\lim_{n\to\infty}\frac{1}{ n^{\frac{1}{n}}} = 0 [/tex]


so while this doe not give me the sum (I'm not excited about that part...), it at least tells me that this series converges. Am I correct?
 

Answers and Replies

  • #2
vela
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Your first step is incorrect. How did you just get rid of the 1/n exponent?
 
  • #3
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Homework Statement



Does it converge, and what is the sum:

[tex] \sum_{n=1}^{\infty}\frac{1}{n n^{\frac{1}{n}}} [/tex]

Homework Equations





The Attempt at a Solution



Please check my method and conclusion:

Using the root test:

[tex] \displaystyle\lim_{n\to\infty}\left|\frac{1}{n n^{\frac{1}{n}}} \right|^{\frac{1}{n}} = \displaystyle\lim_{n\to\infty}\frac{1}{n n^{\frac{1}{n}}} = \displaystyle\lim_{n\to\infty}\frac{1}{n}\displaystyle\lim_{n\to\infty}\frac{1}{ n^{\frac{1}{n}}} = 0 [/tex]
I don't see that you actually took the root. You lost the 1/n exponent right away without changing anything.
so while this doe not give me the sum (I'm not excited about that part...), it at least tells me that this series converges. Am I correct?

You could have simpified things by writing nn1/n as n1+1/n. This looks like a convergent p-series to me.
 
  • #4
Dick
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I don't see that you actually took the root. You lost the 1/n exponent right away without changing anything.


You could have simpified things by writing nn1/n as n1+1/n. This looks like a convergent p-series to me.

NOT a convergent p-series, Mark44. The limit of n^(1/n) as n->infinity is one. Think about a comparison test.
 
  • #5
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NOT a convergent p-series, Mark44. The limit of n^(1/n) as n->infinity is one. Think about a comparison test.
I had a suspicion that it wasn't a p-series, given that the exponent wasn't a constant. Thanks for straightening me out, Dick.
 
  • #6
I had a suspicion that it wasn't a p-series, given that the exponent wasn't a constant. Thanks for straightening me out, Dick.

I'm not sure it matters that the exponent is a constant, as long as it is an integer n>0, because no matter what n is here, the value will be greater than 1. BUT since 1/n converges to 0, then we have the exponent converging to one, and then we have a problem.

I'm still not sure what to do here. As far as comparison goes:

[tex] \sum \frac{1}{nn^{\frac{1}{n}}} \leq \sum \frac{1}{n} [/tex]

But that is not helpful b/c the 1/n series diverges...
 
  • #7
Dick
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I'm not sure it matters that the exponent is a constant, as long as it is an integer n>0, because no matter what n is here, the value will be greater than 1. BUT since 1/n converges to 0, then we have the exponent converging to one, and then we have a problem.

I'm still not sure what to do here. As far as comparison goes:

[tex] \sum \frac{1}{nn^{\frac{1}{n}}} \leq \sum \frac{1}{n} [/tex]

But that is not helpful b/c the 1/n series diverges...

You've got (1/n)*(1/n^(1/n)). The second factor approaches 1. So for large n the second factor is greater than, say, 1/2.
 
  • #8
You've got (1/n)*(1/n^(1/n)). The second factor approaches 1. So for large n the second factor is greater than, say, 1/2.

[tex] \sum \frac{1}{nn^{\frac{1}{n}}} \geq \sum\frac{1}{2n} [/tex]

Since we know that the second series is divergent and smaller than the original, we can say the original must diverge as well.
 
  • #9
Dick
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Best possible outcome, really. Now you don't have to worry about summing it.
 
  • #10
:-) Thanks for the input.
 

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