# Covergence/Divergence of an Infinite Series

1. Mar 18, 2010

### SpringPhysics

1. The problem statement, all variables and given/known data
Does the infinite series $$\sum_{n=1}^{\infty}(\frac{1}{n^(1 + (\frac{1}{n}))})$$ converge?

2. Relevant equations
Power series $$\sum_{n=1}^{\infty}\frac{1}{n^p}$$

3. The attempt at a solution
I used the fact that for a power series, if p>1 the series will converge.
Since $${n^(1+\frac{1}{n})}$$ will always be greater than n, the series will converge.

Is this correct? If not, can someone lead me in the right direction please?

2. Mar 18, 2010

### gabbagabbahey

What does $$\sum_{n=1}^{\infty}\frac{1}{n+1}$$ become when you make the substitution $m=n+1$ (and hence $n=m-1$)?

3. Mar 18, 2010

### SpringPhysics

Sorry if the text isn't clear - the denominator is n^(1 + 1/n).

4. Mar 18, 2010

### gabbagabbahey

In that case, your argument is invalid since the exponent depends on the summation index (It's true that $1+\frac{1}{n}\geq 1$, but it isn't a constant value like the $p$ in your power series convergence test).

You will have to find another convergence test to use instead.

5. Mar 18, 2010

### SpringPhysics

Sorry but do you have any in mind? The only ones I know are the comparison, ratio, and integral tests, and the function makes these tests impractical (at least I think so).

6. Mar 18, 2010

### gabbagabbahey

I think the comparison test is your best bet here.

7. Mar 18, 2010

### SpringPhysics

I thought about that, as in the power series. Since the exponent on n is between 1 and 2, the series sits right in between a diverging (exponent 1) and a converging (exponent 2) series. Wouldn't that be inconclusive?

8. Mar 18, 2010

### SpringPhysics

Sorry for the double post, but if I use the limit comparison test, letting the series {a_n} be the series of the question and {b_n} be 1/n, then the limit as n approaches infinity of a_n/b_n would = 1, but since the series {b_n} does not converge, then {a_n} does not converge as a series either. Is that it?

9. Mar 18, 2010

### Gib Z

If the ratio exists and is non zero, then a_n and b_n converge or diverge together. You limit is 1 (non zero) so your deduction is correct.

10. Mar 18, 2010

### SpringPhysics

Thank you guys so much!