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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:


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:


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:


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 $\forall n\gt0$:


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!

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It's true in the limit, since $\lim_{N\to\infty}\sum_{n=1}^{N}{\frac{1}{n^{2}}} =\lim_{N\to\infty}\int_{{\rm e}^{1/N}}^{\infty}{\left(\frac{1-x^{-N}}{x^{2}-x}\right)\ln{x}\:dx},=\frac{\pi^2‌​}6$ and $1<\frac{\rm e}{2}$. +1 for showing what you've tried. – draks ... Jul 19 '12 at 11:47
up vote 8 down vote accepted

The equation $$\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{2}{n^2{\rm e}}$$ can be rewritten as $$\frac{1}{n^2}=\frac{\rm e}2\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}$$

Now, simply sum this for $n=1,2,\ldots,N$ and get: $$\begin{align}\sum_{n=1}^N\frac{1}{n^2}=&\sum_{n=1}^N\frac{\rm e}2\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}<\sum_{n=1}^N\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\\=&\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}\sum_{n=1}^N{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}\ln{x}\sum_{n=1}^N{\frac{1}{x^{n+1}}\:dx}\end{align}$$

(To get the inequality, we used that the integrand is positive on $(1,\infty)$, which shows that $\int_{{\rm e}^{1/n}}^{\infty}\leq\int_{{\rm e}^{1/N}}^{\infty}$ for $N\geq n$ and equality holds if and only if $N=n$.) The sum here is just the sum of a finite geometric progression. Thus, we have $$\sum_{n=1}^N{\frac{1}{x^{n+1}}}=\sum_{n=1}^N{(x^{-1})^{n+1}}=x^{-2}\sum_{n=1}^{N}{(x^{-1})^{n-1}}=x^{-2}(\frac{(x^{-1})^N-1}{x^{-1}-1})=\frac{x^{-N}-1}{x-x^2}$$

This gives us $$\sum_{n=1}^{N}{\frac{1}{n^{2}}}<\frac{\rm e}{2}\int_{{\rm e}^{1/N}}^{\infty}{\left(\frac{1-x^{-N}}{x^{2}-x}\right)\ln{x}\:dx},$$ which is exactly what we wanted.

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+1 for the elegant proof, thanks! – Shaktal Jul 19 '12 at 12:54

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