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Could anyone help me to find the mistake in the following problem? Based on the formula of the sum of a geometric series: \begin{equation} 1 + x + x^{2} + \cdots + x^{n} + \cdots = \frac{1}{1 - x} \end{equation} \begin{equation} 1 + \frac{1}{x} + \frac{1}{x^{2}} + \cdots + \frac{1}{x^{n}} + \cdots = \frac{1}{1 - 1/x} = \frac{x}{x-1} \end{equation} Adding both equations \begin{equation} 2 + x + \frac{1}{x} + x^{2} + \frac{1}{x^{2}} + \cdots + x^{n} + \frac{1}{x^{n}} + \cdots = \frac{1}{1 - x} + \frac{x}{x-1} = \frac{1-x}{1-x} = 1 \end{equation} So, \begin{equation} 2 + x + \frac{1}{x} + x^{2} + \frac{1}{x^{2}} + \cdots + x^{n} + \frac{1}{x^{n}} + \cdots = 1 \end{equation} And the left side is always bigger than $2$ for $x>0$.

What is wrong?? Thanks in advance

Michael Albanese
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Kikolo
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    I like this! Subtracting $1$ on both sides we get the beautiful identity $$\sum_{n\in \mathbb Z} x^n = 0 $$ – hmakholm left over Monica Dec 27 '14 at 13:52
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    This question reminds me of an extremely embarrassing moment late in my first complex analysis course. Man, I could have proven the Riemann hypothesis with the way I manipulated those Laurent series! – guest Dec 28 '14 at 04:30
  • @HenningMakholm Which you could also prove by noting that $xF(x)=F(x)$ (where $F(x)$ is that… thing). – Akiva Weinberger Oct 14 '15 at 23:58

2 Answers2

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The first series only applies when $|x| < 1$ whereas the second series only applies when $\left|\frac{1}{x}\right| < 1$ (i.e. $|x| > 1$). By adding them, you are assuming that they both apply simultaneously, but they don't (for any $x$).

Ahaan S. Rungta
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Michael Albanese
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The domains of convergence of these two sequences don't coincide. One converges for $|x|>1$ and the other for $|x| < 1$. Therefore, the sum is meaningless.

user141592
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    Actually, only the first series has a radius of convergence. – Did Dec 27 '14 at 00:02
  • "The intervals of convergence of these two series don't coincide." would be a better way of putting it. (Although in one case it's not even an interval in the most conventional sense. Perhaps one could consider it to be the interval "$(1,-1)$" $={x\in\mathbb R : |x|>1}\cup{\infty}$, where this is neither $+\infty$ nor $-\infty$, but is the $\infty$ that is the value of $\tan\frac\pi2$, at both ends of the line. ${}\qquad{}$ – Michael Hardy Dec 27 '14 at 00:48
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    @MichaelHardy: I'd suggest "domains of convergence". – ruakh Dec 27 '14 at 02:09
  • Fixed it :) That will teach me to do math after a glass of wine – user141592 Dec 27 '14 at 02:15
  • @Did: Does the second series have a radius of divergence? – bof Dec 27 '14 at 02:21
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    @bof Definition? – Did Dec 27 '14 at 02:22
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    I would prefer: the domains of convergences have an empty intersection. That they are not identical is not so special, but here one cannot even find a single point where both series converge. – Marc van Leeuwen Dec 27 '14 at 13:49