Name:
Anonymous
2010-02-03 15:45
1+(1+(1+(1+(1+(1+(1+(1+(1+(1+(1+(1+(1+(1+(1+(1+(1+(…)))))))))))))))))=∞
Name:
Anonymous
2010-02-03 18:14
\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}
Name:
Anonymous
2010-02-05 15:19
actually it's div/0, amirite?
Name:
Anonymous
2010-02-05 15:33
1+1+1+1+1+... = -1/2
see zeta function
Name:
Anonymous
2010-02-21 5:37
\begin{array}{c}test \newline test\end{array}
Name:
Someone
2010-03-09 17:04
>>7
I've read something about it, but it wasn't precise. Can you explain a bit moar ?
Name:
Anonymous
2010-03-15 4:22
>>7
FAIL
Keep talking out your ass. It's hilarious.