<s>[INST] $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$; is $p$ divisible by $1997$? 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 $1997$?
 [/INST] 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.
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