Savanah Boone

2022-07-01

Simplifying this logarithm series
$\sum _{i\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}2}^{99}\frac{1}{{\mathrm{log}}_{i}\left(99!\right)}$
How would you evaluate (or at least simplify) this logarithm series?

isscacabby17

Expert

Note that ${\mathrm{log}}_{a}b=\frac{\mathrm{log}b}{\mathrm{log}a}$, for any choice of base of the logarithm. (Mathematicians tend to mean the natural logarithm when they write $\mathrm{log}$.)
In particular, $\frac{1}{{\mathrm{log}}_{a}b}=\frac{\mathrm{log}a}{\mathrm{log}b}={\mathrm{log}}_{b}a$.
$\sum _{i=2}^{99}{\mathrm{log}}_{99!}i={\mathrm{log}}_{99!}\left(\prod _{i=2}^{99}i\right)={\mathrm{log}}_{99!}99!=1$
where in first equality I used the well-known fact that $\sum \mathrm{log}=\mathrm{log}\prod$.