Prove alternating sum equals odd numerator sum with denominators 996-1990

[anemone]

Hi members of the forum,

I have been trying so hard to prove the following:

$$\frac{1989}{2}-\frac{1988}{3}+\frac{1987}{4}-\cdots-\frac{2}{1989}+\frac{1}{1990}=\frac{1}{996}+\frac{3}{997}+\frac{5}{998}+\cdots++\frac{1989}{1990}$$

but to no avail and what I couldn't bear was that my approach gave me a tremendously messy bundle of terms everywhere on the paper...

Could you please at least give me some idea on how to prove it?

Thanks in advance.

 

[Opalg]

Hi members of the forum,

I have been trying so hard to prove the following:

$$\frac{1989}{2}-\frac{1988}{3}+\frac{1987}{4}-\cdots-\frac{2}{1989}+\frac{1}{1990}=\frac{1}{996}+\frac{3}{997}+\frac{5}{998}+\cdots+\frac{1989}{1990}$$

but to no avail and what I couldn't bear was that my approach gave me a tremendously messy bundle of terms everywhere on the paper...

Could you please at least give me some idea on how to prove it?

Thanks in advance.

I would start by converting this expression from arithmetic to algebra. If you put $n=995$ then the equation becomes $$\frac{2n-1}2 - \frac{2n-2}3 +\frac{2n-3}4 -\ldots - \frac2{2n-1} + \frac1{2n} = \frac1{n+1} + \frac3{n+2} + \frac5{n+3} + \ldots + \frac{2n-1}{2n}.$$ The next thing is to ask whether that formula holds for other values of $n$. I checked that it works for $n=2$ and $n=3$, which makes it seem likely to hold in general.

So, how to prove it? Now that you have got rid of all those numbers, it ought to be easier to see what to do next. Take each term on the right side of the formula over to the left (changing the sign, of course) and combine it with the term having the same denominator on the left side. There is still some way to go, but I think you should be able to verify the formula that way. If in doubt, go back to smaller values of $n$, such as $n=4$, and check what happens there.