# Series convergence, root test

1. Feb 3, 2010

### Somefantastik

1. The problem statement, all variables and given/known data

Does it converge, and what is the sum:

$$\sum_{n=1}^{\infty}\frac{1}{n n^{\frac{1}{n}}}$$

2. Relevant equations

3. The attempt at a solution

Please check my method and conclusion:

Using the root test:

$$\displaystyle\lim_{n\to\infty}\left|\frac{1}{n n^{\frac{1}{n}}} \right|^{\frac{1}{n}} = \displaystyle\lim_{n\to\infty}\frac{1}{n n^{\frac{1}{n}}} = \displaystyle\lim_{n\to\infty}\frac{1}{n}\displaystyle\lim_{n\to\infty}\frac{1}{ n^{\frac{1}{n}}} = 0$$

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?

2. Feb 3, 2010

### vela

Staff Emeritus
Your first step is incorrect. How did you just get rid of the 1/n exponent?

3. Feb 3, 2010

### Staff: Mentor

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.

4. Feb 3, 2010

### Dick

NOT a convergent p-series, Mark44. The limit of n^(1/n) as n->infinity is one. Think about a comparison test.

5. Feb 3, 2010

### Staff: Mentor

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. Feb 3, 2010

### Somefantastik

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:

$$\sum \frac{1}{nn^{\frac{1}{n}}} \leq \sum \frac{1}{n}$$

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

7. Feb 3, 2010

### Dick

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. Feb 3, 2010

### Somefantastik

$$\sum \frac{1}{nn^{\frac{1}{n}}} \geq \sum\frac{1}{2n}$$

Since we know that the second series is divergent and smaller than the original, we can say the original must diverge as well.

9. Feb 3, 2010

### Dick

Best possible outcome, really. Now you don't have to worry about summing it.

10. Feb 4, 2010

### Somefantastik

:-) Thanks for the input.