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