$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$; is $p$ divisible by $1997$?

Question

if $p,q\in \mathbb{N}$ and

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$$

why is $p$ divisible by $1997$?

Answer 1

Let $S$ be the LHS. Using $$ 1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{2n}=\frac{1}{n+1}+\cdots+\frac{1}{2n} $$ (which can be proved by induction), we can deduce that $$ S-\frac{1}{1332}=\frac{1}{667}+\cdots+\frac{1}{1331}+\frac{1}{1332}\\ \implies S=\frac{2}{1332}+\frac{1}{667}+\cdots+\frac{1}{1331}=\frac{1}{666}+\cdots+\frac{1}{1331}. $$ This implies $$ S=1997\left(\frac{1}{666\times 1331}+\frac{1}{667\times 1330}+\cdots+\frac{1}{998\times 999}\right). $$ Let the expression in the parentheses equal to $\frac{a}{b}$ where $a$ and $b$ are positive integers satisfying $a<b$ and $\text{gcd}(a,b)=1$. Then because $1997$ is prime, $b$ isn't divisible by $1997$ (the prime factors of $b$ constitute a subset of the prime factors of $\prod_{j=666}^{1331}j$), so we have $1997a|p$ and $b|q$. The claim follows.

Answer 2

Write $$1-\frac 12+\frac 13- \dots+\frac 1{1331}=\left(1+\frac 12+\frac 13+ \dots +\frac 1{1331}\right)-\left(1+\frac 12+ \dots +\frac 1{665}\right)$$

Using $1=2\cdot \frac 12$, and $ \frac 12=2\cdot \frac 14$, and $ \frac 13=2\cdot \frac 16$ etc to eliminate the negative fractions.

Then the sum becomes $$\frac 1{666}+\frac 1{667}+\dots +\frac 1{1331}$$

Now note that $\cfrac 1{n}+\cfrac 1{N-n}=\cfrac N{n(N-n)}$ and apply to $1997=666+1331=667+1330= \dots = 998+999$

Finally $1997$ is a prime, and all the factors in the denominators are $\lt 1997$, so there is no factor which will cancel the $1997$ in the numerator.