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Let $m \geq2$ be an integer. I want to ask how to prove that the sum of the following series is irrational: $$\sum _{n=1}^{\infty} \frac{1}{m^{n^2}}$$

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    What have tried?2012-07-16
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    And what techniques do you know, to prove a number is irrational?2012-07-16
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    @did Indeed, i have no idea of the proof. I will choose the topic of repeating decimal to prove a number is irrational. But I don't know how to work on this series. :(2012-07-16
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    Surely you can write the expansion of this number in base $m$. Does it repeat itself after a while?2012-07-16
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    @did Is that 1/m+ 1/(m^4)+ 1/(m^16)+ 1/(m^25)+........? how to recognize it repeat itself?2012-07-16
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    Think decimals, for example 1/4 in base 3 is 0.0202020... In base m, the digits are 0, 1, ..., m-1, and you want to write down your number in base m.2012-07-16
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    Suppose $m = 10$. What now?2012-07-16
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    @did I can understand what you mean, but m is a variable not a real number. How can I change 1/m in therms of decimals?2012-07-16
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    @Nerd-Herd That is 0.1+0.0001+0.000000001+...2012-07-16
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    Surely $m$ is a fixed integer, otherwise the statement you are trying to prove is false (for example by the inverse function theorem).2012-07-16
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    [link]http://math.stackexchange.com/questions/17310/proving-that-a-series-converges-to-an-irrational-number In fact, he asked the similar questions, but i couldn't understand the answer from Aryabhata.2012-07-16
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    @AlexBecker Oh, I see what you mean. There is a misunderstanding. (Likely due to my ambiguous wording.) yup, although m is a fixed integer, i don't know how to work. :(2012-07-16

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To get a grasp on what is going on here, we start with a simple, seemingly unrelated, problem, namely, we show that the decimal representation of $x=1/7$ is periodic.

We know how to do that, right? One should perform the long division of $1$ by $7$ and very soon the digits produced by the division will begin to cycle. The result is $$x=0.14285714285714285714\ldots,$$ usually abbreviated as $x=0.\overline{142857}$. Thus, the decimal representation is indeed periodic... but now we might want to ask two questions:

  • What causes this periodic expansion?
  • What does it mean?

Well, the meaning is clear: the expansion says that $$x=n\cdot10^{-6}+n\cdot10^{-12}+n\cdot10^{-18}+\cdots$$ with $n=142857$, that is, that $$x=n\cdot10^{-6}\cdot\sum\limits_{i=0}^{+\infty}10^{-6i}=\frac{n\cdot10^{-6}}{1-10^{-6}}=\frac{n}{10^6-1}.$$ In other words, we started from a representation of $x$ as a fraction, namely $x=1/7$, and we reached another representation of $x$ as a fraction, namely $x=142857/(10^6-1)$. If the goal is to compute the decimal representation of $x$, the second fraction is actually quite nice because, due to the simple fact that $142857\lt10^7$, the periodicity of the representation of $x$ becomes obvious. But we still have two mysteries to solve:

  • Why did one obtain $10^6-1$ as the denominator?
  • Why did one obtain $142857$ as the numerator?

At this point, one could note that, since the expression of every rational number as a reduced fraction is unique, one better have $10^6-1=7\cdot n$. In particular, $10^6-1$ should be a multiple of $7$, that is $10^6=1\pmod{7}$. Now, this rings a bell! One knows that, as soon as $k$ and $b$ are relatively prime, $k^{\phi(b)}=1\pmod{b}$. Since $10$ and $7$ are relatively prime and $\phi(7)=6$, indeed $10^6-1$ is a multiple of $7$, and the remark also explains the appearance of $6$ as the exponent of $10$. More importantly, it suggests a reason why the whole shebang holds and, at the same time, a way to vastly generalize our observations to any rational number.

So, we now consider any rational number $x=a/b$. We assume without loss of generality that $a\geqslant1$, $b\geqslant2$ (otherwise $x$ is an integer), $a\leqslant b-1$ (otherwise, shift $x$ by an integer), and that $b$ has no factor $2$ or $5$ (otherwise, multiply $b$ by powers of $5$ or $2$ to get a power of $10$, then multiplying $x$ by this power of $10$ simply shifts the expansion of $x$). Thus, $10$ and $b$ are relatively prime and $10^c=1\pmod{b}$, for some positive integer $c$. This means that $10^c=bd+1$ for some integer $d$, which implies that $x=n/(10^c-1)$ with $n=ad\lt10^c-1$. One gets $x=0.\overline{n_{c-1}n_{c-2}\cdots n_1}$, where $n=\sum\limits_{i=0}^{c-1}n_i\cdot10^i$ with $n_i$ in $\{0,1,\ldots,9\}$ is the decimal representation of $n$. This proves that the decimal representation of $x$ is indeed periodic and, at the same time, yields a way to compute this representation.

(When $a=1$ and $b=7$, indeed one gets $c=6$, $d=999999/7=142857$ and $n=142857$.)


Coming back finally to the question asked, one sees that there is nothing specific to the base $10$ here. Thus, consider any integer $m\geqslant2$ and some real number $x$ in $[0,1]$. Thus, $x=\sum\limits_{i=1}^{+\infty}x_im^{-i}$ for some sequence $(x_i)_{i\geqslant1}$ with values in $\{0,1,\ldots, m-1\}$. Then, $x$ is a rational number if and only if the sequence $(x_i)_{i\geqslant1}$ is ultimately periodic.

In the case at hand, $x_i=1$ when $i$ is a square and $x_i=0$ otherwise, hence the gaps between the digits $1$ are unbounded. This forbids ultimate periodicity, hence $x$ is irrational.