Which is greater: \[\frac{1^{-3}-2^{-3}}{1^{-2}-2^{-2}}-\frac{2^{-3}-3^{-3}}{2^{-2}-3^{-2}}+\frac{3^{-3}-4^{-3}}{3^{-2}-4^{-2}}-\cdots +\frac{2019^{-3}-2020^{-3}}{2019^{-2}-2020^{-2}}\]or \[1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots +\frac{1}{5781}?\]
We have
$\frac{a^{-3}-(a+1)^{-3}}{a^{-2}-(a+1)^{-2}}=\frac{(a+1)^3-a^3}{(a+1)^3a-a^3(a+1)}=\frac{3a^2+3a+1}{a(a+1)(2a+1)}=\frac{1}{a}+\frac{1}{a+1}-\frac{1}{2a+1}$
So first sum is $A=\sum_{a=1}^{2019} (-1)^{a+1} (\frac{1}{a}+\frac{1}{a+1}-\frac{1}{2a+1})=\sum_{a=1}^{2019} (-1)^{a+1} \frac{1}{a}+\sum_{a=1}^{2019} (-1)^{a+1} \frac{1}{a+1}+\sum_{a=1}^{2019} (-1)^{a} \frac{1}{2a+1}=1+\frac{1}{2020}+\sum_{a=1}^{2019} (-1)^{a} \frac{1}{2a+1}=1+\frac{1}{2020}-\frac{1}{3}+\frac{1}{5}-...-\frac{1}{4039}$
Second sum is $B$
$B-A=-\frac{1}{2020}+\frac{1}{4041}-\frac{1}{4043}+\frac{1}{4045}-...+\frac{1}{5781}=-\frac{2021}{2020*4041}-\frac{2}{4041*4035}-\frac{2}{4037*4039}-...-\frac{2}{5779*5781}<0$