Let $\frac{a}{b} = 1 - \frac{1}{2} + \frac13 - \frac14 + ... - \frac{1}{1318} + \frac{1}{1319}$. Show that $1979  | a$. Given that , $a$ and $b$ are coprimes.  I tried to compute in pairs , but I don’t know how to do next.  Any hints or solutions would be appreciated.
The question is $$\text{If }\, p, \ q\in \mathbb{N}, \;1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}.\qquad \text{Prove that } 1979\mid p.$$ So my solution went like this:  $$S=1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}\Rightarrow S=\sum_{k=1}^{1319}\frac{1}{k}-2\sum_{k=1}^{659}\frac{1}{2k}=\sum_{k=660}^{1319}\frac{1}{k}$$ Now this is equivalent to finding the sum of the roots of the equation $$\left(x-\frac{1}{660}\right)\left(x-\frac{1}{661}\right)\dots \left(x-\frac{1}{1319}\right)=0$$. Rewriting this equation, we get $(660x-1)(661x-1)\dotsc(1319x-1)=0$ (we can discard the $\displaystyle \frac{1319!}{659!}$ since $1979$ is a prime). Calculating the sum of the roots, we get the required value to be equal to $(1320+659)\times 330$ which is obviously divisible by $1979$ (again discarding the coeffecient of $x^{660}$).  Now my question is: Using the same method of converting the problem into finding the sum of roots of a polynomial, we get from the original equation something like this: $$(x-1)\left(x+\frac{1}{2}\right)\dots \left(x-\frac{1}{1319}\right)=0 \Rightarrow (x-1)(2x+1)\dotsc(1319x-1)=0$$ So, the sum of roots would be $-1+2-3+4-\dotsc+1318-1319=-660$ But this is not divisible by $1979$. Why does the former solution work while the latter doesn't? Is there a flaw in my reasoning? Note: I know that the question can be solved by some other methods. Please only explain my given doubts and don't instead give another solution  I realised my mistake after some hours of pondering over it. (In the 1st equation) To get the required sum I would need to take calculate $\displaystyle -\frac{\text{coeffecient of }x^{659}\,}{\text{coeffecient of }x^{660}}$. However what I've done is calculate this:  $\displaystyle -\frac{\text{ coeffecient of }x}{\text{coeffecient of }x^{660}}$ which is actually equal to $\text{sum of roots taken }659 \text{ at a time} $. It was just a coincidence that the value I calculated was divisible by $1979$. The correct way, indeed, would be through the link supplied
Is it possible to programmatically stop and start ArcGIS Server map services using Python? As I prefer outer parentheses to grow larger in nested expressions, I happen to insert \bigl and \bigr and its larger cousins a lot. I have always wondered whether there is a way to do this automagically.  Inserting \left and \right prophylacticly before all opening and closing parentheses and hoping that it sorts things out doesn't do the trick in all cases as seen below.  The left column is without any size changing commands. The middle column as a \left \right pair at every parentheses/brace/bracket and the rightmost column shows my preferred version.    As seen in the middle column of the second and third line the \left and \right pair sometimes makes the outer parentheses larger, on the cost of an extra horizontal space between \Pr and (. The extra space is fine in case of really big opening parentheses but here I'd prefer the regular spacing.  \documentclass{scrartcl} \usepackage{amsmath} \usepackage{amsfonts} \begin{document} \begin{align*} &amp;1-(1-F(x))^n&amp;&amp;1-\left(1-F(x)\right)^n&amp;&amp;1-\bigl(1-F(x)\bigr)^n\\ &amp;\Pr(X_{(1)}\le x)&amp;&amp;\Pr\left(X_{(1)}\le x\right)&amp;&amp;\Pr\bigl(X_{(1)}\le x\bigr)\\ &amp;\mathbb{E}[\min\{X_1,X_2\}]&amp;&amp;\mathbb{E}\left[\min\left\{X_1,X_2\right\}\right]&amp;&amp;\mathbb{E}\bigl[\min\left\{X_1,X_2\right\}\bigr]\\ &amp;\left(\pi-\arccos(\frac{y}{r})\right)&amp;&amp;(\pi-\arccos\left(\frac{y}{r}\right))&amp;&amp;\left(\pi-\arccos \left(\frac{y}{r}\right)\right) \end{align*} \end{document}   As mathematical expressions can be arbitrarly complex there is probably no general way to do this. But I’m not asking for a solution of the general case. A partial solution that works in the simple cases shown above would be a big help. 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