Bug in mathematics

>It holds for an infinite case of infinite tests, which is what setting the limit to infinity means.
It just isn't what it means.

in this simple case, the limit has to do with metric, it has to do with [math]\forall[/math].

Saying that x is the limit of a sequence s is saying

[math]\forall\epsilon>0:\exists n\geq 0: |s(n) -x|

[math]
\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} +\frac{1}{8} + ... = a= \frac{1}{2}(1 + \frac{1}{2} + \frac{1}{4} +...) = \frac{1}{2}(1+a) = a
\\\\
\displaystyle
\sum_{n=1}^{\infty}\frac{9}{10^n} = \frac{9}{10} + \frac{9}{100} +\frac{9}{1000} + ... = b = \frac{9}{10}(1 + \frac{9}{10} + \frac{9}{100} +...) = \frac{9}{10}(9+b) = b

[/math]

>It doesnt hold for every case before infinity. It holds for an infinite case of infinite tests, which is what setting the limit to infinity means
No, it isn't.

no one gives Weierstrass the credit for making the concept rigorous.

Wrong.
[eqn] \forall \epsilon > 0 \exists N \geq 0: \forall M\geq N, |s(M) - x| < \epsilon [/eqn]

OMG you're right, I'm retarded sorry

Ignore , this is the right version >you'd think after Calculus V I'd know this shit

[math]
\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} +\frac{1}{8} + ... = a= \frac{1}{2}(1 + \frac{1}{2} + \frac{1}{4} +...) = \frac{1}{2}(1+a) = a \\\\ \displaystyle \sum_{n=1}^{\infty}\frac{9}{10^n} = \frac{9}{10} + \frac{9}{100} +\frac{9}{1000} + ... = b = \frac{9}{10}(9 + \frac{9}{10} + \frac{9}{100} +...) = \frac{9}{10}(1+\frac{b}{9}) = b
[/math]

...

sauce?