Let n = 2006 ! then 1 log 2 n + 1 log 3 n + ? + 1 log 2006 n =

Question

Let $n = 2006!$ then $\frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \dots + \frac{1}{\log_{2006} n} =$

Smart Achievers · Accepted Answer

(4)

$\frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \dots + \frac{1}{\log_{2006} n}$

$= \frac{\log 2}{\log n} + \frac{\log 3}{\log n} + \dots + \frac{\log (2006)}{\log n}$

$= \frac{1}{\log n} [\log 2 + \log 3 + \dots + \log 2006]$

$= \frac{1}{\log n} [\log (1 \cdot 2 \cdot 3 \dots 2006)]$

$= \frac{1}{\log n} \log (2006!) = \frac{1}{\log n} \times \log n = 1$ $(∵ n = 2006!)$

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