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  4. If N = n!, where n is a'fixed integer > 2, then (1/ log2 N) + (1/ log 3 N)+ ...+(1/ log n N)=

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Q. If $N = n!$, where n is a'fixed integer $> 2$, then $\frac{1}{\log_{2} N} + \frac{1}{\log _{3} N}+ ...+\frac{1}{\log _{n} N}= $

COMEDKCOMEDK 2011Probability - Part 2

Solution:

$\frac{1}{\log_{2} N} + \frac{1}{\log _{3} N}+ ...+\frac{1}{\log _{n} N}$
$ = \frac{\log 2}{\log N} + \frac{\log 3}{\log N} + .....+ \frac{\log n}{\log N} $
$ = \frac{1}{\log N} \left(\log 2+\log 3+...+\log n\right)$
$ = \frac{1}{\log N} \log \left(2.3.4.....n\right) $
$ =\frac{1}{\log n!}\log \left(1.2.3.4....n\right) \left(\because \, N = n!\right) $
$ = \frac{\log \left(n!\right)}{\log \left(n!\right)}= 1$