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Source. Example 2.1, preprint pp. 3--4, "cf. Erdős and Straus [6], Theorem 2.26". Read on the rendered pages.

Statement and argument

Example 2.1 (pp. 3--4): $\sum_{n\ge1}\varphi(n)/n!$ and $\sum_{n\ge1}\sigma(n)/n!$ are irrational, $\varphi$ being Euler's totient function and $\sigma(n)$ the sum of the positive divisors of $n$. The paper writes each sum as $1+\sum_{n\ge1}(f(n)-n+1)/n!$, with $f=\varphi$ or $f=\sigma$, and gives a one-line reason for each: $\varphi(p)=p-1$ and $\sigma(p)=p+1$ at primes $p$, with the bounds $0<\varphi(n)\le n-1$ and $n<\sigma(n)=o(n^2)$, which it states for all $n$ (they hold for $n\ge2$); for $\sigma$ it names Lemma 2.1 and (3) as the tools.

The mechanism (Lemma 2.1 and its Remark, p. 3, and formula (3)): with $a_n=n$ and $b_n=\sigma(n)-n+1>0$, a rational sum makes the tails $S_N=(N-1)!\sum_{n\ge N}b_n/n!$ integers for all large $N$, while (3), from $b_n=o(n^2)$, gives $|S_N-b_N/N|<\epsilon$ for large $N$; at a large prime $N=p$, $b_p/p=2/p$, so $0<S_p<1$, which is impossible. For $\varphi$ the same argument runs with $b_n=\varphi(n)-n+1\le0$ for $n\ge2$, $b_p=0$ and the sign reversed. (The identity $\sum_{n\ge1}(n-1)/n!=1$ supplies the rewriting.)

Relation to problem 252

$\sum\sigma(n)/n!$ is the case $k=1$ of problem 252. This example is an elementary reproof; the paper refers to Erdős–Straus 1971, Theorem 2.26, and the rational independence of $1$, $\sum\varphi(n)/n!$ and $\sum\sigma(n)/n!$ is Erdős–Straus 1974, Theorem 3.7 with $a_n=n$. Nothing here concerns $\sigma_k$ for $k\ge2$, where $\sigma_k(n)-n^k+1$ is not $o(n^2)$.

Bears on. #252 (the case $k=1$).