0
$\begingroup$

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?

1 Answers 1

1

For all $ 1\leq n \leq N$ we have $\int^{\infty}_{e^{1/n}} \frac{\log x}{x^{n+1} } dx \leq \int^{\infty}_{e^{1/N}} \frac{\log x}{x^{n+1} } dx.$

Thus the sum on the left is less than this: $\sum_{n=1}^N \left( \int^{\infty}_{e^{1/N}} \frac{\log x}{x^{n+1} } dx \right).$

And now of course, we can exchange finite sums and integrals.