Proving the inequality log(1)1!+log2(2)2!+log3(3)3!+⋯>π4\frac{\log (1)}{1!}+\frac{\log ^2(2)}{2!}+\frac{\log^3(3)}{3!}+\cdots> \frac{\pi }{4}

How to prove this inequality? log(1)1!+log2(2)2!+log3(3)3!+>π4


The left side looks vaguely like the series for exp(x): the terms starting from nth contribute at least as much as the corresponding terms of the series for exp(logn). But for the preceding terms, the inequality goes in the opposite direction.

Answer

As @user2345215 and @Mr.G said in the comments, to answer the question as posed is quite simple.
log(1)1!+log2(2)2!+log3(3)3!+>11n=2logn(n)n!>0.7855>π4

Attribution
Source : Link , Question Author : Community , Answer Author :
Zander

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