# Covergence/Divergence of an Infinite Series

## Homework Statement

Does the infinite series $$\sum_{n=1}^{\infty}(\frac{1}{n^(1 + (\frac{1}{n}))})$$ converge?

## Homework Equations

Power series $$\sum_{n=1}^{\infty}\frac{1}{n^p}$$

## 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?

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

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

gabbagabbahey
Homework Helper
Gold Member
Sorry if the text isn't clear - the denominator is n^(1 + 1/n).

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.

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).

gabbagabbahey
Homework Helper
Gold Member
I think the comparison test is your best bet here.

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?

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?

Gib Z
Homework Helper
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.

Thank you guys so much!