I am doing a question that asks to show that \sum_{n=1}^{N} \frac{1}{n^2} < \frac{e}{2} \int_{e^{1/N}}^{\infty} \left(\frac{1-x^{-N}}{x^2-x}\right)\ln x dx for integer $N>1$.
The proof goes that \sum_{n=1}^N \frac{1}{n^2} = \frac{e}{2}\sum_{n=1}^N \frac{2}{en^2} = \frac{e}{2} \sum_{n=1}^{N} \int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}}dx < \frac{e}{2} \int_{e^{1/N}}^{\infty} \sum_{n=1}^{N} \frac{\ln x}{x^{n+1}} \, dx = \frac{e}{2} \int_{e^{1/N}}^{\infty} \frac{1-x^{-N}}{x^2-x} \ln x dx
However, I am unable to see how \frac{e}{2} \sum_{n=1}^{N} \int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx < \frac{e}{2} \int_{e^{1/N}}^{\infty} \sum_{n=1}^{N} \frac{\ln x}{x^{n+1}}dx. Am I missing something obvious? Also, is there a generality of this kind of integral of summation - summation of integral inequality?