If correlation $n = 2002$, evaluate $\frac{1}{\log_{2}{n!}} + \frac{1}{\log_{3}{n!}} + \frac{1}{\log_{4}{n!}} + ... + \frac{1}{\log_{2002}{n!}}$

Question

Dr. Abubacker · Accepted Answer

(a)
$\frac{1}{\log_{2}{n!}} + \frac{1}{\log_{3}{n!}} + \frac{1}{\log_{4}{n!}} + ... + \frac{1}{\log_{2002}{n!}}$
$= \frac{\log 2}{\log{n!}} + \frac{\log 3}{\log{n!}} + ... + \frac{\log 2002}{\log{n!}}$
$= \frac{\log{(2.3.4\ldots 2002)}}{\log{n!}}$
$= \frac{\log{2002!}}{\log{n!}}$
$= \frac{\log{2002!}}{\log{2002!}} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because n = 2002,$ given]

Q. If correlation $n = 2002$, evaluate $\frac{1}{\log_{2}{n!}} + \frac{1}{\log_{3}{n!}} + \frac{1}{\log_{4}{n!}} + ... + \frac{1}{\log_{2002}{n!}}$

Explanation

Reviewed by Dr. Abubacker

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