Sum of Log

Algebra Level 3

For n ∈ N , n ≥ 2 n\in\mathbb{N},n\geq2 ∑ k = 2 n 1 log ⁡ k n ! = ? \sum_{k=2}^n\dfrac{1}{\log_kn!}=?

Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8. 8! = 1\times2\times3\times\cdots\times8.


The answer is 1.

Zakir Husain by Zakir Husain , 41, India

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1 solution

Tom Engelsman
Apr 9, 2021

Easily solved via a change-of-base in the logarithm:

log ⁡ k n ! = ln ⁡ n ! ln ⁡ k ⇒ Σ k = 2 n ln ⁡ k ln ⁡ n ! ; \large \log_{k} n! = \large \frac{\ln n!}{\ln k} \Rightarrow \large \Sigma_{k=2}^{n} \frac{\ln k}{\ln n!};

or 1 ln ⁡ n ! ⋅ Σ k = 2 n ln ⁡ k \frac{1}{\ln n!} \cdot \Sigma_{k=2}^{n} \ln k ;

or 1 ln ⁡ n ! ⋅ ( ln ⁡ 2 + ln ⁡ 3 + . . . + ln ⁡ n ) ; \frac{1}{\ln n!} \cdot (\ln 2 + \ln 3 + ... + \ln n);

or 1 ln ⁡ n ! ⋅ ln ⁡ ( 2 ⋅ 3 ⋅ . . . ⋅ n ) ; \frac{1}{\ln n!} \cdot \ln (2 \cdot 3 \cdot ... \cdot n);

or 1 ln ⁡ n ! ⋅ ln ⁡ n ! = 1 . \frac{1}{\ln n!} \cdot \ln n! = \boxed{1}.

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