Series convergence / divergence

[goraemon]

Homework Statement

Does the following series converge or diverge? If it converges, does it converge absolutely or conditionally?
\sum^{\infty}_{1}(-1)^{n+1}*(1-n^{1/n})

Homework Equations

Alternating series test

The Attempt at a Solution

I started out by taking the limit of $a_n: Lim_{n\rightarrow\infty}(1-n^{1/n})$=1-1=0.
So via the alternating series test, the original series converges.

Next, I have to figure out whether it converges absolutely or conditionally, and this is where I'm stuck. I suppose I have to first find out whether the term $a_n$, which is $(1-n^{1/n})$, diverges or converges. But what test do I use for this? The limit test is silent because as I found above, the limit is zero. I've tried ratio test and limit comparison test to no avail. Root test doesn't seem to lead anywhere. It's not a geometric or telescoping series so those options are out. Maybe I could use comparison test, but I don't know what term I would use for comparison. Integral test seems really difficult. So I'm stuck and would appreciate any helpful pointers.

 

[lurflurf]

think about [(n^(1/n)-1)/(1/n)]/n
compare to log(n)/n

 

[goraemon]

Wait, so...I gather you're saying that $\sum(-1)^{n+1}*(1-n^{1/n})=\sum(-1)^n*(n^{1/n}-1)$, so when we consider the positive term $(n^{1/n}-1)$, it can be re-written as:

$\frac{(n^{1/n}-1)*n}{n}$, aka:


$\frac{\frac{(n^{1/n}-1)}{\frac{1}{n}}}{n}$?

And from here we do a comparison test with $\frac{\ln n}{n}$? OK, From punching some numbers into the calculator, I can see that $\frac{\frac{(n^{1/n}-1)}{\frac{1}{n}}}{n} > \frac{\ln n}{n}$, so via the comparison test the positive term diverges, meaning the original alternating series converges conditionally?

But if so, then I'm not sure how to show rigorously that $\frac{\frac{(n^{1/n}-1)}{\frac{1}{n}}}{n} > \frac{\ln n}{n}$, other than to say it seems to be true for some arbitrarily large n...am I on the right track here?

 

[lurflurf]

^Exactly
n^(1/n)-1>=0 for all n (n=1,2,3,4,...)
x>=log(1+x) for all x (x>=0)
thus
n^(1/n)-1>log{1+[n^(1/n)-1]}=log(n)/n
∑log(n)/n diverges so too does ∑[n^(1/n)-1]

 

[scurty]

But if so, then I'm not sure how to show rigorously that $\frac{\frac{(n^{1/n}-1)}{\frac{1}{n}}}{n} > \frac{\ln n}{n}$, other than to say it seems to be true for some arbitrarily large n...am I on the right track here?

You can use derivatives to show this (increasing/decreasing functions).