# If $g\geq2$ is an integer, then $\sum\limits_{n=0}^{\infty} \frac{1}{g^{n^{2}}}$ and $\sum\limits_{n=0}^{\infty} \frac{1}{g^{n!}}$ are irrational

How do we show that if $g \geq 2$ is an integer, then the two series $$\sum\limits_{n=0}^{\infty} \frac{1}{g^{n^{2}}} \quad \ \text{and} \ \sum\limits_{n=0}^{\infty} \frac{1}{g^{n!}}$$ both converge to irrational numbers.

Well, i tried to see what happens, if they converge to a rational but couldn't get anything out it.

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Both are transcendental numbers. This follows from Roth's theorem on the rate of approximations to irrational numbers by rationals. See also this question. The key concept is "irrationality measure", see for example this entry from MathWorld.

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I would upvote, but I am out of votes! –  Aryabhata Jan 12 '11 at 23:29
@Moron: Just wait 30 minutes-you'll have more. –  Ross Millikan Jan 12 '11 at 23:30
@Ross: Yeah, I know. Just commenting so that I remember and can come back to this later from my activity tab :-) –  Aryabhata Jan 12 '11 at 23:36
Dear Andres, At least for the second number, probably Liouvill'e's original Diophantine approximation result is enough. (I mention this just because it is a lot easier than Roth's theorem.) Best wishes, –  Matt E Jan 13 '11 at 16:47
@Matt E : Oh, sure. The mention of Roth's theorem was specifically to address the first series. –  Andres Caicedo Jan 13 '11 at 16:59

The numbers are irrational, as they are infinite non-repeated "decimal" numbers in base-$g$.

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ALthough that seems intuitively true, I only remember that repeating decimals implied rationality but have not seen how to show non-repeating decimals implies irrationality –  Arjang Jan 13 '11 at 5:35
@Arjang: This is a well known fact. If $(b,g) = 1$, then for some $m$ we have $g^m = 1 \mod b$. This gives $g^m \times \frac{a}{b} = (bk+1) \times \frac{a}{b} = ka + \frac{a}{b}$, which implies a repeating representation for $\frac{a}{b}$. –  Aryabhata Jan 13 '11 at 6:29

Just for the record: the usual proof of irrationality is as follows (say for the first sum). Suppose that the sum is $p/q$. Multiply by $qg^{n^2}$, where $n$ is "big enough". We get that $$\sum_{m > n} \frac{q}{g^{m^2-n^2}}$$ is an integer. However, the latter is bounded by the geometric series $$q \sum_{t > 0} g^{-Nt} = \frac{q}{g^N} \cdot \frac{1}{1-g^{-N}} = \frac{q}{g^N-1},$$ where $N = (n+1)^2 - n^2 = 2n+1$. When $n$ is big enough, $q/(g^N-1) < 1$, contradiction.

This proofs works for any series $$\sum_{n=1}^\infty \frac{1}{\prod_{i=1}^n a_i}$$ where the non-zero integers $a_i$ satisfy $|a_i| \rightarrow \infty$.

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