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Prove that if $\{n_k\}$ is a strictly increasing sequence of positive integers, then the sum of the series $$\sum_{k=1}^{\infty} \frac{2^{n_k}}{(n_{k})!}$$ is an irrational number.

This is just a problem, given by a friend of mine, when we were doing some problems in Analysis. And if I remember correctly, he mentioned a theorem of "Legendre", to be used for proving this result. But this looks quite interesting.

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What exactly is the question? Do you just want a solution to the problem, a reference for relevant work of Legendre, or what? – Jonas Meyer Aug 27 '10 at 2:21
What is the question? Certainly one can analyze the convergence of this series by using the ratio test. – J. M. Aug 27 '10 at 2:23
@Jonas Meyer: Yes, i want to prove that this sum is an irrational number. – anonymous Aug 27 '10 at 2:26
There is a lemma that might be relevant: the largest power of 2 that divides (x_k)! is x_k minus the sum of the digits of x_k in base 2. – Qiaochu Yuan Aug 27 '10 at 2:33
You might find your answer by looking at Liouville's theorem on algebraic and Liouville numbers: – Eric Haengel Aug 27 '10 at 16:05
up vote 1 down vote accepted

I received this solution via email. I am TeXing it out here.

We can write $n!= 2^{\alpha(n)}\beta(n)$, where $\beta(n)$ is odd. From a theorem of Legendre we have $\alpha(n)=n -\gamma(n)$ where $\gamma(n)$ denotes the sum of ones in the binary representation of $n$. Moreover, have $$ \sum\limits_{k=1}^{\infty} \frac{2^{n_k}}{n_{k}!} = \sum\limits_{n=1}^{\infty} \delta_{n} \frac{2^n}{n!}$$ where $\delta_{n}=1$ if $n=n_{k}$ and $\delta_{n}=0$, otherwise. Suppose, contrary to our claim, if $$\sum\limits_{n=1}^{\infty} \delta_{n} \frac{2^n}{n!}= \frac{p}{q}$$ Write $q=2^{s}t$, where $t$ is odd. Take $N=2^{r} > \max \{t,2^{s+2}\}$, it then follows that $\frac{\beta(n)}{n} \in \mathbb{N}$. Therefore $2^{s}\beta(N)\frac{p}{q}=\frac{\beta(N)}{t} p \in \mathbb{N}$. A simple calculation shows that $N!=2^{N-1}\beta(N)$. Multiplying the equality $$ \frac{p}{q} = \sum\limits_{n=1}^{N} \delta_{n} \frac{2^n}{n!} + \sum\limits_{n=N+1}^{\infty} \delta_{n} \frac{2^n}{n!}$$ by $2^{s}\beta(N)$, we get $$ 2^{s}\beta(N)\frac{p}{q} = 2^{s}\beta(N) \sum\limits_{n=1}^{N} \delta_{n} \frac{2^n}{n!} + 2^{s}\beta(N) \sum\limits_{n=N+1}^{\infty} \delta_{n} \frac{2^n}{n!} \quad (\star) $$ Note that $$ 2^{s} \beta(N) \sum\limits_{n=1}^{N} \delta_{n} \frac{2^n}{n!} = 2^{s} \sum\limits_{n=}^{N} \delta_{n} \frac{\beta(N)2^{n}}{2^{\alpha(n)}\beta(n)}$$ Since $\beta(n)$ divides $\beta(N)$ we see that the first term on the right hand side of $(\star)$ is an integer. To obtain contradiction we will show that, $$0 < 2^{s}\beta(N) \sum\limits_{n=N+1}^{\infty} \delta_{n} \frac{2^{n}}{n!} <1$$ Indeed, we have $$2^{s}\beta(N) \sum\limits_{N+1}^{\infty} \delta_{n} \frac{2^n}{n!} = 2^{s-N+1}N! \sum\limits_{n=N+1}^{\infty} \frac{2^n}{n!}$$ $$= 2^{s+2} \sum\limits_{n=N+1}^{\infty} \delta_{n}2^{n-N-1} \frac{N!}{n!} < \frac{2^{s+2}}{N+1} \sum\limits_{n=N+1}^{\infty} \Bigl(\frac{2}{N+2}\Bigr)^{n-N-1}$$ $$=\frac{2^{s+2}}{N+1} \frac{N+2}{N} < \frac{2^{s+3}}{N+1} < 1$$

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In this self-answer I prove that $e^2$ is irrational, and the method generalizes easily to any infinite subsequence of the series.

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