Ratio & Root Tests: Convergence & Divergence

[zonk]

Homework Statement

Test for convergence and divergence.

Homework Equations

$\sum_{i=1}^{n} \frac{n!}{3^n}$

$\sum_{i=1}^{n} \frac{1}{{(log (n))}^{1/n}}$

The Attempt at a Solution

For the first one I get $\frac{(n+1)}{3}$ After applying the ratios. Taking the limit as n approaches infinity, it is greater than 1, so it should diverge. But the book says it converges.

For the second one, I think we have to take the (1/n)th root, which converts it to $\sum_{i=1}^{n} \frac{1}{(log (n))}$. Is this correct resoning?

 

[zonk]

Never mind the first one. I was reading the wrong answers.

 

[nickalh]

For the second one,
sum $\frac{1}{{log (n)}^{1/n}}$
we've already got an nth root in the denominator.
It's tempting to take the nth root, because it's similar to nth root of $x^n$.
Taking the nth root of an nth root isn't what we want.Different, Simpler Method:
What is
$lim {(log (n))}^{1/n}$ as i->oo
?Nitpicky note:
Technically, as given each term is a constant.
The entire sum would evaluate to $\frac{n}{{(log (n))}^{1/n}}$
because a constant term added n times, is n * constant.
Don't you mean
$\sum_{i=1}^{oo} \frac{1}{{(log (i))}^{1/i}}$ ?
I've assumed that's what you mean throughout my post.

Rule of thumb: with the oo above the sum, it doesn't matter if the variable is i or n, just be consistent that the variable underneath the sum and the variable in the term is the same.
It becomes less nitpicky and more important to distinguish, when you start doing radius of convergence.

Finally, please let us know if you can work it from here, or have further questions.