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I am trying some question in sequence and series of an institute in which I don't study and I was unable to solve this particular question .

Question: Find the sum :$\sum_{l=2}^{100} \frac{1} {\log_{l} 100!}$ .

A 0.01

B 0.1

C 1

D 10

I am sorry but I would not be able to provide an attempt on this because I am absolutely clueless On how to aproach this even though I am good in Logarithms and sequence and series.

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    $\begingroup$ $\frac{1} {log_{l} 100!}=\log_{100!}l$; can you continue now? $\endgroup$
    – Conrad
    Nov 13, 2020 at 18:21
  • $\begingroup$ @Conrad +1 : so elegant that it deserves to be an answer. $\endgroup$ Nov 13, 2020 at 20:16

3 Answers 3

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$\log_i 100!=\frac{\log_{100!}100!}{\log_{100!}i}=\frac{1}{\log_{100!} i}$. So $$\sum_{i=2}^{100} \frac{1}{\log_i 100!}=\sum_{i=2}^{100}\log_{100!} i= \log_{100!}\prod_{i=2}^{100} i=\log_{100!} 100!=1.$$

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$$S=\sum_{k=2}^{100} \frac{1}{\log_k100!}=\sum_{k=2}^{100} \frac{\log k}{\log 100!}=\frac{\log 2+\log 3+\log 4+\log 5+....+\log 100}{\log100!}$$ $$=\frac{\log (2.3.4....100)}{\log 100!}=\frac{\log 100!}{\log 100!}=1.$$ Option (C) is correct.

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$$\sum_{l=2}^{100} \frac{1} {\log_{l} 100!}=\sum_{l=2}^{100} \log_{100!}{l}=\log_{100!}{2}+\log_{100!}{3}+\log_{100!}{4}+...+\log_{100!}{100}=\log_{100!}({2\times3\times4\times...\times100})=\log_{100!}{100!}=1$$ .

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