sum of natural logs

Algebra Level pending

k = 2 784965315 ln ( k ) ln ( 784965315 ! ) = ? \sum_{k=2}^{784965315}\ln(k) - \ln(784965315!)= \ ?

Notation: ! ! denotes the factorial notation.

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1 solution

Tommy Li
May 17, 2020

( k = 2 784965315 ln ( k ) ) ln 784965315 ! = ( ln 2 + ln 3 + ln 4 + + ln 784965315 ) ln 784965315 ! = ln ( 2 × 3 × 4 × × 784965315 ) ln 784965315 ! = ln 784965315 ! ln 784965315 ! = 0 \large \displaystyle \left( \sum_{k=2}^{784965315} \ln{(k)} \right) -\ln{784965315!} \\ = (\ln{2} +\ln{3}+\ln{4} +\cdots + \ln{784965315} ) -\ln{784965315!} \\ = \ln{(2 \times 3 \times4 \times \cdots \times 784965315) } -\ln{784965315!} \\ = \ln{784965315!} - \ln{784965315!} \\ = 0

Probably should make it obvious that the sigma applies only to ln(k), not ln(784965315!) as well.

Justin Park - 1 year ago

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