What about this one?

What about this one?

Other urls found in this thread:

en.wikipedia.org/wiki/Harmonic_series_(mathematics)
en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Comparison_test
sites.google.com/site/scienceandmathguide/
latex.codecogs.com/eqneditor/editor.php
sciweavers.org/free-online-latex-equation-editor
twitter.com/SFWRedditVideos

Looks nice. Well presented. Clear font. Good work, OP.

Thank you, mate!

>1/2nlogn which blows up.

How do I prove that it is divergent?

[math] n! < n^n [/math]
[math] \sum_{n=2}^\infty \frac{1}{n \log(n)} [/math] diverges.

Hey bruv, try posting in one single thread for your questions. Posting them here

will give you more luck on getting an answer, have a good one.

ratio test

Are we already at this part of Calc II this year?

You need to watch some videos and do some reading; if you don't understand this than you're in for a huge god damned problem once you get past this basic bitch shit.

Stirling approximation.

[math]
\displaystyle
\sum_{n=2}^{ \infty}\frac{1}{ \text{ln} \; n!}
[/math]

is log(n) the most J U S T function in existance

1/(n^1.0001) converges but 1/n log(n) doesn't

wouldn't that be convergent, as n gets larger the denominator gets larger and larger and the numerator remains the same, so it converges to some finite value.
then again i'm a brainlet and suck at this calc shit

[math]
\displaystyle \sum_{n=1}^{ \infty} \frac{1}{n}
[/math]
isn't convergent either,
takes a bit of work to sort them out

>isn't convergent either,
What?
Let [math] y = \displaystyle \sum_{n=1}^{ \infty} \frac{1}{n} \implies y + \displaystyle \sum_{n=1}^{ \infty} (-1)^n \frac{1}{n} = \displaystyle \sum_{n=1}^{ \infty} \frac{1}{2n} = \frac{1}{2}y [/math]

But this sum I have added converges to zero so this yields [math] y = \frac{1}{2}y [/math] which implies [math] y = 0 [/math]. Therefore

[eqn] \displaystyle \sum_{n=1}^{ \infty} \frac{1}{n} = 0 [/eqn]

This is a trivial result any retard should be able to prove. I am sad to see someone so retarded like you posting here. Please go back to /b/.

Your implies is wrong.
[math]\infty=\frac{1}{2}\infty[/math] too.

[math]\infty=\frac{1}{2}\infty[/math]

en.wikipedia.org/wiki/Harmonic_series_(mathematics)

>bad latex
>wikipedia page he just edited to make it read what he wanted it to read

Kek these are some top tier arguments. I guess I have been destroyed!

>I'm trolling, not retarded!
Sure, bud.

leave spaces

Stupid preview kept lying to me.
[math]\infty = \frac{1} {2} \infty[/math]

en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Comparison_test

>Proof by wikipedia
Nice, do you think I can try this for my upcoming midterms? Just link a wikipedia article on the test like you are doing here. How many marks do you think I'll get?

[math] n! \leq n^n \implies \log (n!) \leq \log (n^n) \implies \frac{1}{\log (n!)} \geq \frac{1}{\log (n^n)} = \frac{1}{n\log (n)} [/math]
Therefore, if we can show that [math] \sum_{n=2}{\infty} \frac{1}{n\log (n)} [/math] diverges, then [math] \sum_{n=2}{\infty} \frac{1}{\log (n!)} [/math] diverges.
Notice that [math] \frac{1}{n\log (n)} [/math] is strictly decreasing and positive. This allows the integral test.
[math] \int_{2}^{\infty} \frac{1}{t\log (t)} dt \stackrel{x=e^u}{=} \int_{\log (2)}^{\infty} \frac{1}{u} du = + \infty [/math]
Therefore the series diverges.

write the proof on a piece of paper then

2+2=4 no matter what the media

The fastest way is still to use Stirling formula,
ln (n!) ~ n ln n
therefore it diverges, by comparison to bertrand series

TIL
1 + 1/2 + 1/3 + ... = 0

ty mathboy

[eqn] \sum_{n=1}^\infty (-1)^n \frac{1}{n} = \log(2) [/eqn]

[math]- \log(2) [/math] actually.

Exactly.
>log
Could you please explain what you mean by this? What is this "log". Please express it as a rational function. I mean, you wouldn't be so stupid as to think that transcendental functions exist right?

[eqn] \log(x) = \int_1^x \frac{1}{t} dt [/eqn]

exp and ln are perfectly defined as either the limit of a power series or as the solution of a differential equation which is equivalent to saying

I said rational function. Anyone can draw a bunch of weird lines and claim they have a function.

You start adding them and see what direction it goes in.

>∞

How do I make them symbols?

Read the sticky.

[Catalog] >> Veeky Forumsguide
>> sites.google.com/site/scienceandmathguide/
>> Veeky Forums LaTeX Tutorial

for testing, try at
latex.codecogs.com/eqneditor/editor.php
sciweavers.org/free-online-latex-equation-editor

right-click on pretty formulas
to see the code that produced it

protip: put a space before each backslash

Whaddabout a verb in your sentence, illiterfag?

...

L0Lno fgt pls retreat into obscurity now