- #1

end3r7

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## Homework Statement

(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}}\Bigg)[/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}}}\Bigg)[/tex]

(f) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(1 - e^{\frac{-1}{n}}\Bigg)[/tex]

## Homework 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

## 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) I am not sure which test to apply

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