If $n = 20!$, evaluate the sum: $\frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \dot

Question

If $n = 20!$, evaluate the sum: $\frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \dots + \frac{1}{\log_{20} n}$.

Accepted Answer

$1$ Explanation: Using the change-of-base property of logarithms, we know that $\frac{1}{\log_a b} = \log_b a$.

Applying this property to each term of the series, we get:
$\log_n 2 + \log_n 3 + \log_n 4 + \dots + \log_n 20$.

Using the product rule for logarithms, $\log_b x + \log_b y = \log_b(xy)$:
$= \log_n (2 \cdot 3 \cdot 4 \dots 20) = \log_n (20!) = \log_n n$.

Since $\log_n n = 1$, the sum is equal to $1$.

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