leave spaces
What about this one?
Caleb Walker
Anthony Edwards
Stupid preview kept lying to me.
[math]\infty = \frac{1} {2} \infty[/math]
Caleb Wright
>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?
Andrew Sanders
[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.
Christopher Wright
write the proof on a piece of paper then
2+2=4 no matter what the media
Jackson Myers
The fastest way is still to use Stirling formula,
ln (n!) ~ n ln n
therefore it diverges, by comparison to bertrand series
Juan Lopez
TIL
1 + 1/2 + 1/3 + ... = 0
ty mathboy
Caleb Reyes
[eqn] \sum_{n=1}^\infty (-1)^n \frac{1}{n} = \log(2) [/eqn]
Dylan Cox
[math]- \log(2) [/math] actually.