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}

Question 1

How to prove this inequality? $\frac{\log (1)}{1!}+\frac{\log ^2(2)}{2!}+\frac{\log^3(3)}{3!}+\cdots> \frac{\pi }{4}$

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

Question 2

As @user2345215 and @Mr.G said in the comments, to answer the question as posed is quite simple.
$\frac{\log (1)}{1!}+\frac{\log ^2(2)}{2!}+\frac{\log^3(3)}{3!}+\cdots > \sum_{n=2}^{11} \frac{\log^n(n)}{n!}>0.7855> \frac{\pi }{4}$

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}

Answer

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