Question

If n = 1983 !, compute the sum
$\frac{1}{lo{g}_{2}n}+\frac{1}{lo{g}_{3}n}+\frac{1}{lo{g}_{4}n}+\cdots \frac{1}{lo{g}_{1983}n}.$

Answer 1

$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+\cdots +\frac{1}{{\log }_{1983}n}=\frac{1}{\frac{\log n}{\log 2}}+\frac{1}{\frac{\log n}{\log 3}}+\frac{1}{\frac{\log n}{\log 4}}+\cdots +\frac{1}{\frac{\log n}{\log 1983}}=\frac{\log 2}{\log n}+\frac{\log 3}{\log n}+\frac{\log 4}{\log n}+\cdots +\frac{\log 1983}{\log n}=\frac{\log 2+\log 3+\log 4+\cdots +\log 1983}{\log n}=\frac{\log (2\times 3\times 4\times \cdots \times 1983)}{\log n}=\frac{\log (1983!)}{\log n}=\frac{\log n}{\log n}=1$