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

1. Feb 7, 2008

### end3r7

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

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
$$\frac{\frac{1}{n}}{\frac{1}{n^{1 + \frac{1}{n}}}} = n^{n}$$
That limit is 1. And since its absolute value converges (a) converges.

(b) Basic Comparion test wtih <= $$\frac{e}{n^{2}}$$

(c) Alternating series test

(e) Integral test

(f) Im not sure which test to apply

Last edited: Feb 7, 2008
2. Feb 7, 2008

### end3r7

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

Is (d) divergent?

3. Feb 7, 2008

### 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

4. Feb 7, 2008

### 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.