The following question was in an entrance exam:
Show that, if $n\gt0$, then: $\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{2}{n^2{\rm e}}$ You are allowed to assume $\lim_{x\to\infty}{\frac{\ln{x}}{x}}=0$. Hence explain why, if $1\lt a\lt b$, then: $\int_{b}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}\lt\int_{a}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}$
Deduce that: $\sum_{n=1}^{N}{\frac{1}{n^{2}}}<\frac{\rm e}{2}\int_{{\rm e}^{1/N}}^{\infty}{\left(\frac{1-x^{-N}}{x^{2}-x}\right)\ln{x}\:dx},$ Where $N\in\mathbb{N}:N\gt1$.
The first part I believe I integrate by parts, such that:
$\int{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{1}{n}\int{x^{-n-1}\:dx}-\frac{x^{-n}\ln{x}}{n}$
Clearly $\int{x^{-n-1}\:dx}=-\frac{x^{-n}}{n}$, so we have:
$\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\left.\frac{x^{-n}(n\ln{x}+1)}{n^{2}}\right|_{{\rm e}^{1/n}}^{\infty}=\frac{{\rm e}^{-\frac{n}{n}}(\frac{n}{n}+1)}{n^{2}}-0=\frac{2}{n^{2}{\rm e}}$
As required. To prove the next inequality, all that is required is to demonstrate that $\left.\frac{\ln{x}}{x^{n+1}}\right|_{x=1}\geq0$, $\lim_{x\to\infty}{\frac{\ln{x}}{x^{n+1}}}\geq0$, and $\left(\frac{\ln{x}}{x^{n+1}}\right)'\neq0$, $\forall x\in[1,\infty)$ and $\forall n\gt0$.
The first inequality is verified simply by observing that $\ln{1}=0$, therefore $\frac{\ln{1}}{1^{n+1}}=0$, $\forall n$.The second inequality can be written as:
$\lim_{x\to\infty}{\left(\frac{1}{x^{n}}\frac{\ln{x}}{x}\right)},$
And as we know $\lim_{x\to\infty}{\frac{\ln{x}}{x}}=0$, and $\lim_{x\to\infty}{\frac{1}{x^{n}}}=0$, the second inequality must also be true. To verify the second inequality we simply differentiate using the quotient rule and look for critical points:
$\frac{d}{dx}{\left(\frac{\ln{x}}{x^{n+1}}\right)}=\frac{x^{n}-(n+1)x^{n}\ln{x}}{x^{2n+2}}=\frac{1-(n+1)\ln{x}}{x^{n+2}}$
As $\ln{x}$ is a monotonically increasing function, and at $x=1$, $\ln{x}=0$, we can show that $\forall n\gt 0$, and $x\gt1$, $\left(\frac{\ln{x}}{x^{n+1}}\right)\leq 0$, therefore, the function is positive for all values of $x\gt 0$, which means by the fundamental theorem of calculus that if $1\lt a\lt b$, then $\forall n\gt0$:
$\int_{b}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}<\int_{a}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}$
As required.
However, I am stuck for the final part of the question. My first thought was that as $\frac{1}{n^{2}}$ is monotonically decreasing, $\forall n\gt0$; any integral performed over the region $(0,\infty)$ will have a positive error term, therefore, we can replace the summation with $\int_{1}^{N}{\frac{1}{n^2}\:dn}=\left.-\frac{1}{n}\right|_{1}^{N}=-\frac{1}{N}+1$. I am not sure how to perform the second integral, however.
Thanks in advance!