Simplifying this logarithm series

Question

$$\sum_{i\; =\; 2}^{99}{\frac{1}{\log _{i}\left( 99! \right)}}$$

How would you evaluate (or at least simplify) this logarithm series?

Answer 1

Note that $\log_a b = \frac{\log b}{\log a}$, for any choice of base of the logarithm. (Mathematicians tend to mean the natural logarithm when they write $\log$.)

In particular, $\frac1{\log_a b} = \frac{\log a}{\log b} = \log_b a$.

Therefore your sum is:

$$ \sum_{i=2}^{99} \log_{99!} i = \log_{99!} \left( \prod_{i=2}^{99} i \right) = \log_{99!} 99! = 1 $$

where in first equality I used the well-known fact that $\sum \log = \log \prod$.