Question

Here's another easy one:
SIMPLIFY
\[\Large \sum_{k=2}^n \frac{1}{\log_{k} u}\]
where u is an arbitrary valid number.

Answer 1

im thinking:\[\sum_{k=2}^{n}\frac{1}{\log_k u}=\sum_{k=2}^{n}\frac{\log_k k}{\log_k u}=\sum_{k=2}^{n} \log_u k=\log_u (n!)\]Not sure if anything else can be done though.