Question

This thing...
1+x+x^2+x^3+...=1/(1-x)
Is true for |x| < 1 right?

Answer 1

If so why is Euler (in step 8) allowed to do that?
http://math.ucr.edu/home/baez/qg-winter2004/zeta.pdf

Answer 2

This is known as an infinite series. Hold on i'll check the link.

Answer 3

http://www.youtube.com/watch?v=w-I6XTVZXww

Answer 4

Yeah that's where I got it from... there's a linked vid where he does a better proof. In it he differentiates the above, then plugs in x = -1, which isn't in the domain...

Answer 5

Not sure then.

Answer 6

Haha, thanks anyway!

Answer 7

@ParthKohli any thoughts?

Answer 8

You are right about that.
Even though I read the document, I'm sorry to say that I have not reached that level of mathematics.

Answer 9

much truth.
very amaze
wow

Answer 10

The document accepts that the series diverges, but it is using the equation from step 7.

Answer 11

doge much smart

Answer 12

Here's that video I'm talking about
http://www.youtube.com/watch?v=E-d9mgo8FGk
In it he says it's true for x<1

Answer 13

is that chinese

Answer 14

i know it is math but what kind of math is THAT!!!

Answer 15

I've watched only one proof. Didn't know there was a second one too!

Answer 16

\[s(n) =1+x+x^2+x^3+...+x ^{n-1}\]
\[x \times s(n) =1+x+x^2+x^3+...+x ^{n-1} =x+x^2+x^3+...+x ^{n-1}+x ^{n1}\]
now
s(n) -x s(n) =\[x-s(n) =1 -x ^{n} \rightarrow s(n)=\frac{1 -x ^{n} }{1-x }\]

\[\lim_{n \rightarrow \infty} s(n)=\lim_{n \rightarrow \infty}\frac{1 -x ^{n} }{1-x }=\frac{1 }{1-x } \rightarrow \lim_{n \rightarrow \infty}x^n=0 \rightarrow |x|<1 \]

Answer 17

"By using the equation in part 7 to evaluate the formal power series"

formal power series exist independently of considerations of convergence

Answer 18

anyways the Abel sum is simple:$$A\sum_{n=1}^\infty (-1)^{n+1} n=
\lim_{t\to 1^-}\sum_{n=1}^\infty (-1)^{n+1}nt^n=-\lim_{t\to 1^-}\sum_{n=1}^\infty n(-t)^n$$

Answer 19

now we know:$$\sum_{n=1}^\infty x^n=\frac{x}{1-x},|x|<1\\\frac{d}{dx}\sum_{n=1}^\infty x^n=\frac{d}{dx}\frac{x}{1-x}\\\sum_{n=1}^\infty\frac{d}{dx}x^n=\frac1{(1-x)^2}\\\sum_{n=1}^\infty nx^{n-1}=\frac1{(1-x)^2}\\\sum_{n=0}^\infty(n+1)x^n=\frac1{(1-x)^2}\\\sum_{n=0}^\infty nx^n+\sum_{n=0}^\infty x^n=\frac1{(1-x)^2}\\\sum_{n=0}^\infty nx^n=\frac1{(1-x)^2}-\frac1{1-x}$$ergo we find:$$-\lim_{t\to1^-}\left(\frac1{(1+t)^2}-\frac1{1+t}\right)=\frac14$$

Answer 20

thus \(\displaystyle A\sum_{n=1}^\infty(-1)^{n+1}n=\frac14\)

Answer 21

note in the sum we used \(x\mapsto -t\)