Define the integral $I_{n}$ as follows for $n$ an integer greater than $1$: $I_{n}:=\int_{1}^{e}\frac{\ln x}{x^n}dx$
Is it true that
$I_{n}\leq \frac{1}{n-1}\left(1-\frac{1}{e^{n-1}}\right)?$
Define the integral $I_{n}$ as follows for $n$ an integer greater than $1$: $I_{n}:=\int_{1}^{e}\frac{\ln x}{x^n}dx$
Is it true that
$I_{n}\leq \frac{1}{n-1}\left(1-\frac{1}{e^{n-1}}\right)?$
Hint: See Davide Giraudo's answer to your last question.
$ \int_1^e{1\over x^n}\,dx= {x^{-n+1}\over -n+1}\biggl|_1^e={e^{-n+1}\over -n+1} -{1\over -n+1}={1\over n-1}(1-e^{-n+1}). $
Now use $ 0\le\ln x\le 1$ for $1\le x\le e$ to obtain $ \int_1^e{\ln x\over x^n}\,dx\le \int_1^e{1\over x^n}\,dx={1\over n-1}(1-e^{-n+1}). $