How to prove that $ 1 + x + x^2 + x^3 + \cdots = \frac 1{1-x}$ where $|x| < 1$?
I guess I fail to see how one can prove this. Sorry if duplicate or perhaps there is a website with how to prove results for the not well-behaved infinite sequences.
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Sedumjoy
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1I proved this in the first several lines in my solution: https://math.stackexchange.com/questions/2382965/what-does-sum-n-0-infty-cos1n-converge-to/2382991#2382991 Maybe you can work out the details and check your solution against what I wrote. – Mee Seong Im Aug 05 '17 at 23:57
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cancellation helps... – Aug 06 '17 at 00:00
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1@Sedumjoy You can show the equality for the above using the method in the above link or use MacLaurin series (see below). – Mee Seong Im Aug 06 '17 at 00:04
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Possible duplicate of Value of $\sum\limits_n x^n$ – Simply Beautiful Art Aug 06 '17 at 00:54
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Using \begin{equation} 1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x} \end{equation} for all $x$ and all natural $n$ and $\lim_{n\to\infty}x^n=0$ for $|x|<1$ yields the desired equality.
Frieder Jäckel
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It comes from the concept of Taylor/Maclaurin series and is based on the formula
$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a) \, (x-a)^n}{n!}$$
where $f^{(n)}(a)$ is the $n$th derivative of $f$ at $a$. Evaluate this for $f(x) = \frac{1}{1-x}$ and $a=0$ and you will have your answer.
It’s worth noticing that this sum looks like it would diverge because each term is of a higher degree than the previous. However, for fractional $x$, each term actually gets smaller, so the series converges.
gen-ℤ ready to perish
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1The Taylor series does tell that one has for some neighborhood of $x=0$, $\sum x^n=\frac{1}{1-x}$. How do you show that the series is valid for all $x$ with $|x|<1$? – Aug 06 '17 at 01:31
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@Jack Off the top of my head and without trying it on paper, I suppose one could use the ratio test to determine the interval of convergence. – gen-ℤ ready to perish Aug 06 '17 at 03:05