TITLE: Summing reciprocal logs of different bases
QUESTION [6 upvotes]: I recently took a math test that had the following problem:
$$
\frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!}
$$
The sum is equal to 1. I understand that the logs can be broken down into (first fraction shown)
$$
\frac{1}{\log_{2}1 + \log_{2}2 + \log_{2}3 + \dots + \log_{2}50}
$$
How do the fractions with such irrational values become $1$? Is there a formula or does one simply need to combine fractions and use the basic properties of logs?

REPLY [5 votes]: This is a general result.  We can write for any $N$
$$\begin{align}
\sum_{n=2}^N\frac{1}{\sum_{m=2}^N \log_n(m)}&=\sum_{n=2}^N\frac{1}{\sum_{m=2}^N \frac{\log_b (m)}{\log_b(n)}}\\\\
&=\frac{\sum_{n=2}^N\log_b(n)}{\sum_{m=2}^N\log_b(m)}\\\\
&=1
\end{align}$$
where we used $\log_n(m)=\frac{\log_b(n)}{\log_b(m)}$.  And we are done!