# Trouble with tests of divergence... Help me see the bigger picture!

• Sep 16th 2013, 06:42 AM
DrKittenPaws
Trouble with tests of divergence... Help me see the bigger picture!
I am having trouble with a couple of convergence/divergence tests for series. So far in class we've talked about the following:
• nth-term test
• geometric series test
• p-series test
• integral test
• comparison test
• ratio test
• root test

I've been working on harder problems but I am stuck. Could you point me towards the right direction and help me see the bigger picture?

$\displaystyle \sum_{n=1}^{\infty }\frac{n+1}{n^3+1}$
$\displaystyle f(x)=\frac{x+1}{x^3+1}$

$\displaystyle \frac{d}{dx}(x+1)< \frac{d}{dx}(x^3+1) \therefore f(x) \textup{ is decreasing.}$
Since the denominator grows faster than the numerator, then that implies that it diverges. According to Wolfram, the ratio and root test would be inconclusive but it would converge with the comparison test. So does it diverge or converge? What would I compare it to with the comparison test?

$\displaystyle \sum_{n=1}^{\infty }\frac{n!}{n^n}$
$\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=\lim_{n\rightarrow \infty}\frac{(n+1)n!}{(n+1)^{n+1}}=\lim_{n\rightar row \infty}(\frac{n}{n+1})^n$

What do I do after that? According to Wolfram the $\displaystyle \lim_{n\rightarrow \infty}(\frac{n}{n+1})^n=\frac{1}{e}<1 \therefore$ it converges. Why is it equal $\displaystyle \frac{1}{e}$?

$\displaystyle \sum_{n=1}^{\infty }3^{\frac{1}{n}}$
I have no idea how to even start this... Any suggestions?

$\displaystyle \sum_{n=1}^{\infty }\frac{1}{ln(n)^{n}}$
$\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=\lim_{n\rightarrow \infty}\frac{ln(n)^{n}}{ln(n+1)^{n+1}}=?$
How do I even begin to break that down so things start cancelling?

$\displaystyle \frac{1}{2}+\frac{1}{2}\cdot \frac{4}{4}+\frac{1}{2}\cdot \frac{4}{4}\cdot \frac{7}{6}+\cdots +\frac{1\cdot 4\cdot 7\cdots \cdots (3n-2)}{2\cdot 4\cdot 6\cdots \cdots (2n)}+\cdots$
I have no idea how to begin this either!

I know I have a lot of questions but I would greatly appreciate any help you give me. I think I'm missing a key part. Thank you so much!
• Sep 16th 2013, 07:45 AM
DrKittenPaws
Re: Trouble with tests of divergence... Help me see the bigger picture!
EDIT:

1) I decided to go with limit comparison test and got an answer so solved!

2) still unsolved

3) Used divergence test so solved!

4) Still unsolved

5) Still unsolved
• Sep 16th 2013, 08:12 AM
Plato
Re: Trouble with tests of divergence... Help me see the bigger picture!
Quote:

Originally Posted by DrKittenPaws
I am having trouble with a couple of convergence/divergence tests for series. So far in class we've talked about the following:
• nth-term test
• geometric series test
• p-series test
• integral test
• comparison test
• ratio test
• root test

$\displaystyle \sum_{n=1}^{\infty }\frac{n+1}{n^3+1}$

It is best to do a direct comparison here:
$\displaystyle \frac{n+1}{n^3+1}\le\frac{n+n}{n^3}=\frac{2}{n^2}$ which is a p-series.

Quote:

Originally Posted by DrKittenPaws
$\displaystyle \sum_{n=1}^{\infty }\frac{n!}{n^n}$
$\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=\lim_{n\rightarrow \infty}\frac{(n+1)n!}{(n+1)^{n+1}}=\lim_{n\rightar row \infty}(\frac{n}{n+1})^n$
Why is it equal $\displaystyle \frac{1}{e}$?

You must learn this basic fact: $\displaystyle {\lim _{n \to \infty }}{\left( {1 + \frac{a}{{bn + c}}} \right)^d} = {e^{\frac{{ad}}{b}}}$

Example: Because $\displaystyle {\left( {\frac{{n - 1}}{{n + 1}}} \right)^n} = {\left( {1 + \frac{{ - 2}}{{n + 1}}} \right)^n}$ we know that
$\displaystyle {\lim _{n \to \infty }}{\left( {\frac{{n - 1}}{{n + 1}}} \right)^n} = {e^{ - 2}}$.

You will find yourself needing that kind of manipulation over and over again.

As for $\displaystyle \sum\limits_n {{3^{\frac{1}{n}}}}$ use the nth term test $\displaystyle {\lim _{n \to \infty }}\sqrt[n]{3} = ?$
• Sep 16th 2013, 08:22 AM
DrKittenPaws
Re: Trouble with tests of divergence... Help me see the bigger picture!
Quote:

Originally Posted by Plato
It is best to do a direct comparison here:
$\displaystyle \frac{n+1}{n^3+1}\le\frac{n+n}{n^3}=\frac{2}{n^2}$ which is a p-series.

You must learn this basic fact: $\displaystyle {\lim _{n \to \infty }}{\left( {1 + \frac{a}{{bn + c}}} \right)^d} = {e^{\frac{{ad}}{b}}}$

Example: Because $\displaystyle {\left( {\frac{{n - 1}}{{n + 1}}} \right)^n} = {\left( {1 + \frac{{ - 2}}{{n + 1}}} \right)^n}$ we know that
$\displaystyle {\lim _{n \to \infty }}{\left( {\frac{{n - 1}}{{n + 1}}} \right)^n} = {e^{ - 2}}$.

You will find yourself needing that kind of manipulation over and over again.

As for $\displaystyle \sum\limits_n {{3^{\frac{1}{n}}}}$ use the nth term test $\displaystyle {\lim _{n \to \infty }}\sqrt[n]{3} = ?$

I did a LCT on #1 by letting bn = 1/n2 and it worked out, too. Thanks!

Thank you for explaining #2.

Would you mind helping me with the rest?
• Sep 16th 2013, 08:32 AM
Plato
Re: Trouble with tests of divergence... Help me see the bigger picture!
Quote:

Originally Posted by DrKittenPaws
Would you mind helping me with the rest?

What would the root test do on $\displaystyle \sum\limits_{n > 1} {\frac{1}{{{{\ln }^n}(n)}}}~?$