I used similar technique as Fourier's proof of irrationality of $e$ https://en.wikipedia.org/wiki/Proof_that_e_is_irrational to show that this series is indeed an irrational number but I was wondering if there are other elementary proofs, specially techniques related to polynomials with rational coefficients. Any hint, complete proof, reference, etc. is much appreciated.

EDIT : presenting my proof as @AndréNicolas requested : I just followed the same steps; let's assume the series is $s$. Now choose $k_0$ such that $m^{k_0^2} > q$. Define $x = q m^{k_0^2} \left(s - \sum_{k=1}^{k_0}{\frac{1}{m^{k^2}}} \right)$. Now if $s$ is not irrational then $s=\frac{p}{q}$ for some non-negative $p$ and $q$. Then it is easy to check that $x$ is an integer. Now we need to reach to a contradiction by proving that $x < 1$. After simple manipulations, $x = q m^{k_0^2} \sum_{k=k_0+1}^{k_0}{\frac{1}{m^{k^2}}}$ Now, for $k=k_0+1,k_0+2, \ldots$, we have $k^2 > k+k_0^2$ so $$ q m^{k_0^2} \sum_{k=k_0+1}^{k_0}{\frac{1}{m^{k^2}}} < q m^{k_0^2} \sum_{k=k_0+1}^{k_0}{\frac{1}{m^{k+k_0^2}}} = \frac{q}{m^{k_0^2+1}}\frac{1}{1-\frac{1}{m}}=\frac{1}{m-1}$$ which follows from our assumption about $q$ and $k_0$.

  • $\begingroup$ can't you apply Liouville's theorem to show it is transcendental ? $\endgroup$
    – mercio
    Aug 21, 2015 at 19:01
  • $\begingroup$ Thanks. There are some little typos, but it will work. $\endgroup$ Aug 21, 2015 at 22:09
  • $\begingroup$ @mercio: it would be great if you can help me to show that $s$ can't be a root of some polynomial with rational coefficients $\endgroup$
    – Ali
    Aug 23, 2015 at 17:28

1 Answer 1


If $m$ is an integer greater than $1$, one can show that the base $m$ expansion is not ultimately periodic, because it has strings of $0$'s of increasing length.


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