Are these the best tests for convergence of the following series?

end3r7

1. The problem statement, all variables and given/known data
(a) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{(-1)^{n}}{n^{1 + \frac{1}{n}}}\Bigg)[/tex]
(b) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{e^{\frac{1}{n}}}{n^{2}}\Bi gg)[/tex]
(c) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{(-1)^{n}n!}{n^{n}}\Bigg)[/tex]
(d) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{1}{n^{1 + \frac{1}{n}}}\Bigg)[/tex]
(e) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{ln(x)}{n^{\frac{3}{2}}}\Bi gg)[/tex]
(f) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(1 - e^{\frac{-1}{n}}\Bigg)[/tex]


2. Relevant equations
The test that we have messed with are:
Telescoping, Geometric, P-Series, Ratio, Root, Simple Comparison, Limit Comparison, Absolute Convergence, Alternating Series, Dirichlet, Integral, Gauss


3. The attempt at a solution

First, I got they all converge:

(a) (d)
I worked 'd' first. I did a limit comparison test with 1/n
[tex]\frac{\frac{1}{n}}{\frac{1}{n^{1 + \frac{1}{n}}}} = n^{n}[/tex]
That limit is 1. And since its absolute value converges (a) converges.

(b) Basic Comparion test wtih <= [tex]\frac{e}{n^{2}}[/tex]

(c) Alternating series test

(e) Integral test

(f) Im not sure which test to apply

end3r7

(d) is wrong,a dn so is (a), not sure how to prove if they are convergent (if they are).

Is (d) divergent?

end3r7

(a) is convergent by alternating series test.

Did I end up proving (d) divergent or is my proof wrong?

I'm also quite lost on (f) since I can't integrate that function nicely

Dick

Too many questions at once! d) is divergent. It's the same as 1/(n*n^(1/n)). For n^(1/n) show that the log of that approaches zero. So n^(1/n) approaches 1. So you can do a comparison with say 1/(2n). For f) expand e^(-1/n) in a power series using e^x=1+x+x^2/2!+etc and keep only the terms that matter.