# convergent series, sequences?

I want to construct a sequence of rational numbers whose sum converges to an irrational number and whose sum of absolute values converges to 1.

I can find/construct plenty of examples that has one or the other property, but I am having trouble find/construct one that has both these properties.

Any hints(not solutions)?

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Find two irrational numbers $a > 0$ and $b < 0$ such that $a-b = 1$ but $a+b$ is irrational. Create a series with positive terms that sum to $a$ and another series with negative terms that sum to $b$. Combine the two series.

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would n't combining the terms of the two series, suffer from the issues one encounters rearranging the terms in a series which causes changes to the sum of the series. –  Hardy Oct 27 '12 at 20:29
The series will be absolutely convergent and can be rearranged at will. –  mrf Oct 27 '12 at 20:32
that's a good point. Thanks –  Hardy Oct 27 '12 at 20:36
Great, it worked out! Tanks! –  susan Oct 27 '12 at 21:22

Here is a probabilistic way. Consider an i.i.d. sequence $(X_n)_{n\geqslant1}$ of Bernoulli random variables such that $\mathbb P(X_n=+1)=\mathbb P(X_n=-1)=\frac12$. Then $X=\sum\limits_{n\geqslant1}2^{-n}X_n$ is uniformly distributed on $[-1,1]$ hence $X$ is irrational with probability $1$. And naturally, $\sum\limits_{n\geqslant1}2^{-n}|X_n|=1$.

A deterministic way is as follows. Pick any irrational number $z$ in $(0,1)$, with binary expansion $z=\sum\limits_{n\geqslant1}z_n2^{-n}$, where each $z_n$ is in $\{0,1\}$. For every $n\geqslant1$, consider $x_n=(2z_n-1)2^{-n}$. Then $\sum\limits_{n\geqslant1}x_n=2z-1$ is irrational and $|2z_n-1|=1$ for every $n\geqslant1$ hence $\sum\limits_{n\geqslant1}|x_n|=1$.

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This seems unnecessarily complicated to me, but I find it very interesting that it's your first response! –  Benjamin Dickman Oct 27 '12 at 20:29
Thanks! Both nice solutions! –  susan Oct 27 '12 at 21:23

Let $\alpha$ be an irrational number in $(0,1)$. Then for suitable choice of $\epsilon_n\in\{\pm1\}$, you can achieve $$\sum_{n=1}^\infty \frac{\epsilon_n}{2^n}=\alpha\qquad\text{and}\qquad\sum_{n=0}^\infty \frac{1}{2^n}=1.$$ To do so, define $\epsilon_n$ recursively: If $\epsilon_k$ is already known for $k<n$ such that $|s_{n-1}-\alpha|<\frac{1}{2^{n-1}}$ for the partial sum $s_{n-1}=\sum_{k=1}^{n-1}\frac{\epsilon_k}{2^k}$, select $\epsilon_n=+1$ if $\alpha>s_{n-1}$ and $\epsilon_n=-1$ if $\alpha<s_{n-1}$. The case $\alpha=s_{n-1}$ does not occur, because $\alpha$ is irrational. With this choice, $|s_n-\alpha|=|s_{n-1}+\frac{\epsilon_n}{2^n}-\alpha|= |\frac1{2^n}-\frac{\alpha-s_{n-1}}{\epsilon_n}|<\frac 1{2^n}$ because $<\frac{\alpha-s_{n-1}}{\epsilon_n}<2\cdot\frac 1{2^n}$.

Another way to look at this is: The $\frac{1-\epsilon_n}2\in\{0,1\}$ are essentially the binary expansion of $\frac{1-\alpha}2$.

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