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∞

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 2010-02-03 19:16

\sum_{n=0}^{\infty} n

Name: Anonymous 2010-02-03 19:49

orly?

Name: Anonymous 2010-02-04 6:14

1+1+...=∞

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 2010-02-11 12:49

http://www.iwantapal.com/

Name: Anonymous 2010-02-12 16:27

Law of Consistency. :/

Name: Anonymous 2010-02-12 16:31

'sup Peano

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 2010-03-09 19:59

>>12
moar
www.usingenglish.com

Name: Anonymous 2010-03-15 4:22

>>7
FAIL

Keep talking out your ass. It's hilarious.

∞

1 Name: Anonymous 2010-02-03 15:45

2 Name: Anonymous 2010-02-03 18:14

3 Name: Anonymous 2010-02-03 19:16

4 Name: Anonymous 2010-02-03 19:49

5 Name: Anonymous 2010-02-04 6:14

6 Name: Anonymous 2010-02-05 15:19

7 Name: Anonymous 2010-02-05 15:33

8 Name: Anonymous 2010-02-11 12:49

9 Name: Anonymous 2010-02-12 16:27

10 Name: Anonymous 2010-02-12 16:31

11 Name: Anonymous 2010-02-21 5:37

12 Name: Someone 2010-03-09 17:04

13 Name: Anonymous 2010-03-09 19:59

14 Name: Anonymous 2010-03-15 4:22