# $\sum_{n=1}^{N} \int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx < \int_{e^{1/N}}^{\infty} \sum_{n=1}^{N} \frac{\ln x}{x^{n+1}}dx$

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?

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## 1 Answer

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.

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