# What is wrong with the sum of these two series?

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

What is wrong?? Thanks in advance

• I like this! Subtracting $1$ on both sides we get the beautiful identity $$\sum_{n\in \mathbb Z} x^n = 0$$ Dec 27 '14 at 13:52
• 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! Dec 28 '14 at 4:30
• @HenningMakholm Which you could also prove by noting that $xF(x)=F(x)$ (where $F(x)$ is that… thing). Oct 14 '15 at 23:58

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 $x$).

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.

• Actually, only the first series has a radius of convergence.
– Did
Dec 27 '14 at 0:02
• "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. ${}\qquad{}$ Dec 27 '14 at 0:48
• @MichaelHardy: I'd suggest "domains of convergence". Dec 27 '14 at 2:09
• @bof Definition?
– Did
Dec 27 '14 at 2:22
• 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. Dec 27 '14 at 13:49