Absolute Convergence, Conditional Convergence or divergence

[Asphyxiated]

Absolute Convergence, Conditional Convergence or divergence...

Homework Statement

$$\sum_{n=1}^{\infty} \frac {(-2)^{n}}{n^{n}}$$

Homework Equations

$$\lim_{n \rightarrow \infty} | \frac {a_{n+1}}{a_n}| < 1 \;\; absolute\; convergence$$

$$\lim_{n \rightarrow \infty} | \frac {a_{n+1}}{a_n}| > 1 \;\; divergence$$

$$\lim_{n \rightarrow \infty} | \frac {a_{n+1}}{a_n}| = 1 \;\; inconclusive$$

Other tests may be used such as AST but this section is about using the ratio test and we are suppose to use that unless it is inconclusive.

The Attempt at a Solution

First I changed the sum by factoring out the alternating part of it, so maybe I went wrong there:


$$\sum_{n=1}^{\infty} \frac {(-2)^{n}}{n^{n}} = \sum_{n=1}^{\infty} \frac {(-1)^{n}(2)^{n}}{n^{n}}$$

So if that's right then here is what I did to set up the ratio test:

$$\lim_{n \rightarrow \infty} | \frac {(-1)^{n+1}(2)^{n+1}}{(n+1)^{n+1}}* \frac {n^{n}}{(-1)^{n}(2)^{n}}|$$

$$\lim_{n \rightarrow \infty} | \frac {(-1)(2)(n^{n})}{(n+1)^{n+1}}|$$

At this point I drop the (-1) due to the absolute value bars, the denominator can be written as (n+1)^n(n+1) due to the exponent but I don't know where to go from here:

$$\lim_{n \rightarrow \infty} | \frac {2n^{n}}{(n+1)^{n}(n+1)}|$$

This is evidently equal to zero if I type it into my TI-89 which makes it absolutely convergent but I can't see how to get there algebraically. If I try to use L'Hopital's rule in the numerator I will get n^n (ln(n+1)) for as many time as I feel like taking the derivative. And if I type the fraction into maple and tell it to simply I simply get the same function back again.

Thanks for the help in advance!

 

[LCKurtz]

Homework Statement

$$\sum_{n=1}^{\infty} \frac {(-2)^{n}}{n^{n}}$$

Homework Equations

$$\lim_{n \rightarrow \infty} | \frac {a_{n+1}}{a_n}| < 1 \;\; absolute\; convergence$$

$$\lim_{n \rightarrow \infty} | \frac {a_{n+1}}{a_n}| > 1 \;\; divergence$$

$$\lim_{n \rightarrow \infty} | \frac {a_{n+1}}{a_n}| = 1 \;\; inconclusive$$

Other tests may be used such as AST but this section is about using the ratio test and we are suppose to use that unless it is inconclusive.

The Attempt at a Solution

First I changed the sum by factoring out the alternating part of it, so maybe I went wrong there:


$$\sum_{n=1}^{\infty} \frac {(-2)^{n}}{n^{n}} = \sum_{n=1}^{\infty} \frac {(-1)^{n}(2)^{n}}{n^{n}}$$

So if that's right then here is what I did to set up the ratio test:

$$\lim_{n \rightarrow \infty} | \frac {(-1)^{n+1}(2)^{n+1}}{(n+1)^{n+1}}* \frac {n^{n}}{(-1)^{n}(2)^{n}}|$$

$$\lim_{n \rightarrow \infty} | \frac {(-1)(2)(n^{n})}{(n+1)^{n+1}}|$$

At this point I drop the (-1) due to the absolute value bars, the denominator can be written as (n+1)^n(n+1) due to the exponent but I don't know where to go from here:

$$\lim_{n \rightarrow \infty} | \frac {2n^{n}}{(n+1)^{n}(n+1)}|$$

This is evidently equal to zero if I type it into my TI-89 which makes it absolutely convergent but I can't see how to get there algebraically. If I try to use L'Hopital's rule in the numerator I will get n^n (ln(n+1)) for as many time as I feel like taking the derivative. And if I type the fraction into maple and tell it to simply I simply get the same function back again.

Thanks for the help in advance!

Try writing it as

$$2\left( \frac n {n+1}\right)^n\cdot \frac 1 {n+1}$$

and observe that n/(n+1) < 1.

 

[Asphyxiated]

Hey thanks man! I need to get better and writing these expressions in many not-immediately-obvious ways.

 

[Gib Z]

Comparison test works well here: $$\frac{ 2^n}{n^n} = \left( \frac{2}{n} \right)^n \leq \left( \frac{2}{4} \right)^n$$ for $$n \geq 4$$

 

[Asphyxiated]

Hey Gib Z:

I didn't use that because we skipped it in my class. I know nothing of the comparison test.