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I'm just starting to read p-adic Numbers, An Introduction by Fernando Q. Gouvêa. The first chapter is rather informal, but Fact 1.3.1 says,

For each integer $M>0$ there exists an integer $n$ such that $\sum_{k=1}^n \frac{2^k}{k}$ is divisible by $2^M$.

This is supposed to be an example of a fact that can be proved fairly easily with $p$-adic numbers that would be difficult or impossible to prove without them, but I don't believe that it's true, unless "divisible by $2^M$" means something other than "equal to an integral multiple of $2^M$." Indeed, let $p$ be the largest prime $\le n$. Then $n<2p$, for otherwise, by Bertrand's postulate, there is a prime $q$ with $p<q<2p\le n$, in contradiction of the definition of $p$. Then subtract $\frac {2^p}{p}$ from both sides; on the left-hand side, none of the denominators is divisible by $p$, so the sum is a fraction, in lowest terms, whose denominator is not divisible by $p$, whereas the denominator on the right-hand side is $p$. (I'm assuming that $M$ is sufficiently large that $p \ne 2$.)

Only a very informal sketch of the proof of this "fact" is given, so I can't critique it. The notes at the back of the book cite Problem 10.10 and its solution in Exercises in Number Theory by D.P. Parent, but I don't have access to this book.

Do I misunderstand the statement, or is my argument above incorrect? Or is the statement in the book in error? If the latter, how should it be changed?

Thanks for any assistance you can give me.

Brian M. Scott
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saulspatz
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    I'm guessing divisibility here means that $2^M$ divides the numerator of the sum. In other words, expressing the sum as a product of distinct primes, $2^M$ divides the product of primes with positive exponents. – Mathmo123 Oct 08 '16 at 21:29
  • @BrianM.Scott Thanks for the edit. – saulspatz Oct 08 '16 at 22:07
  • @Mathmo123 Thanks. That certainly sounds more reasonable, but I need to think about it. Do you happen to know if that's easy to prove with p-adic numbers? Gouvêa says that the infinite sum converges to 0, p-adically. – saulspatz Oct 08 '16 at 22:10
  • @Mathmo123, your guess is right on the mark. The statement is purely $2$-adic, and ignores other primes than $2$. – Lubin Oct 09 '16 at 04:31
  • Gouvêa’s outline is about as good a $p$-adic proof as can be found, I think. The key to it is that, unlike the complex situation, the $p$-adic logarithm is a homomorphism from the multiplicative group consisting of its domain of convergence, to the additive group of the field. It must therefore vanish on all roots of unity that are in its domain of convergence. – Lubin Oct 10 '16 at 02:46
  • @Lubin Thanks you. – saulspatz Oct 10 '16 at 14:58
  • Related: https://math.stackexchange.com/q/4543367/96384 and https://math.stackexchange.com/q/1990667/96384. – Torsten Schoeneberg Nov 30 '22 at 02:12

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