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Let $p = 6n+1$ be a prime number. Prove that the numerator of the rational number $$ \sum_{k=1}^{4n}\frac{(-1)^{k-1}}{k} $$ is divisible by $p$.

What I've tried: Let \begin{align} g(x) &= (x+1)(2-x)(x+3)(4-x) \cdots (4n-x) \\ &= a_{4n} x^{4n} + a_{4n-1} x^{4n-1} + \cdots + a_1 x + a_0 \end{align} then the unreduced form of the numerator of the alternating harmonic sum is then $a_1$. Since the sum only goes up to $-\frac{1}{4n}$ the question is equivalent to showing $p = 6n+1\mid a_1$. From here I tried showing $g'(0)$ is divisible by $p$ but didn't get anywhere.

Note: This problem is from Andrej Dujella's Number Theory. It's an exercise from the second chapter divisibility, so there might be a solution without mods.

Sammy Black
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llin
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1 Answers1

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I don't offhand see how to use what you've done. Instead, using Proof of $\sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}$, with $n = k$ and $N = 2n$ in the linked question, gives that

$$S = \sum_{k=1}^{4n}\frac{(-1)^{k-1}}{k} = \sum_{k=1}^{2n}\frac{1}{2n + k} = \sum_{k=2n+1}^{4n}\frac{1}{k}$$

Note the denominator of the last term is $4n$, which is $p = 6n + 1$ minus the denominator of the first term, i.e., $2n + 1$. This also occurs with the second and second-last terms, etc. Thus, pairing the terms in this way results in

$$\begin{equation}\begin{aligned} S & = \sum_{k=2n+1}^{3n}\left(\frac{1}{k}+\frac{1}{p-k}\right) \\ & = \sum_{k=2n+1}^{3n}\frac{(p-k)+k}{k(p-k)} \\ & = \sum_{k=2n+1}^{3n}\frac{p}{k(p-k)} \\ & = p\left(\sum_{k=2n+1}^{3n}\frac{1}{k(p-k)}\right) \end{aligned}\end{equation}$$

In lowest terms, the denominator of the summation result above is a factor of $\operatorname{lcm}(2n + 1, 2n + 2, \ldots, 4n)$, so all of its prime factors are smaller than $p$. This means the numerator of the rational number $S$ has a factor of $p$.

As you suggested, the solution (such as shown above) doesn't necessarily require the use of mods.

John Omielan
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