# Integral-Summation inequality.

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.

• 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
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.