Answer: (a) \(\displaystyle \sum_{r=1}^{N} \frac{\ln r}{r^{2}}\lt \frac{2+3\ln 2}{4}-\frac{1+\ln N}{N}.\)

(b) A lower bound is \(\displaystyle \sum_{r=1}^{N} \frac{\ln r}{r^{2}}\gt \frac{\ln 2+1}{2}-\frac{\ln N+1}{N}+\frac{\ln N}{N^{2}}.\)

Let \(f(x)=\dfrac{\ln x}{x^2}\) for \(x\ge 2\).

Since

\(\displaystyle f'(x)=\frac{1-2\ln x}{x^3},\)

and \(\ln x\ge \ln 2\gt \tfrac12\) for \(x\ge 2\), it follows that \(f'(x)\lt 0\). So \(f\) is decreasing for \(x\ge 2\).

(a)

First,

\(\displaystyle \sum_{r=1}^{N}\frac{\ln r}{r^2}=\frac{\ln 2}{4}+\sum_{r=3}^{N}\frac{\ln r}{r^2},\)

because the \(r=1\) term is \(0\).

As \(f\) is decreasing, the rectangles of width 1 with heights \(f(3),f(4),\dots,f(N)\) lie below the curve on \([2,3],[3,4],\dots,[N-1,N]\). Hence

\(\displaystyle \sum_{r=3}^{N}\frac{\ln r}{r^2}\lt \int_{2}^{N}\frac{\ln x}{x^2}\,dx.\)

Therefore

\(\displaystyle \sum_{r=1}^{N}\frac{ln r}{r^2}\lt \frac{\ln 2}{4}+\int_{2}^{N}\frac{\ln x}{x^2}\,dx.\)

Now

\(\displaystyle \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x},\)

so

\(\displaystyle \int_{2}^{N}\frac{\ln x}{x^2}\,dx=\left[-\frac{\ln x+1}{x}\right]_{2}^{N}=-\frac{\ln N+1}{N}+\frac{\ln 2+1}{2}.\)

Substituting,

\(\displaystyle \sum_{r=1}^{N}\frac{\ln r}{r^2}\lt \frac{\ln 2}{4}-\frac{\ln N+1}{N}+\frac{\ln 2+1}{2}.\)

Combine the constants:

\(\displaystyle \frac{\ln 2}{4}+\frac{\ln 2+1}{2}=\frac{2+3\ln 2}{4}.\)

Hence

\(\displaystyle \sum_{r=1}^{N}\frac{\ln r}{r^2}\lt \frac{2+3\ln 2}{4}-\frac{1+\ln N}{N}.\)

(b)

For a lower bound, use the rectangles with heights \(f(2),f(3),\dots,f(N-1)\). Since \(f\) is decreasing, these rectangles lie above the curve on \([2,N]\). Thus

\(\displaystyle \sum_{r=2}^{N-1}\frac{\ln r}{r^2}\gt \int_{2}^{N}\frac{\ln x}{x^2}\,dx.\)

Also,

\(\displaystyle \sum_{r=1}^{N}\frac{\ln r}{r^2}=\sum_{r=2}^{N-1}\frac{\ln r}{r^2}+\frac{\ln N}{N^2}.\)

So

\(\displaystyle \sum_{r=1}^{N}\frac{\ln r}{r^2}\gt \int_{2}^{N}\frac{\ln x}{x^2}\,dx+\frac{\ln N}{N^2}.\)

Using the value of the integral,

\(\displaystyle \sum_{r=1}^{N}\frac{\ln r}{r^2}\gt \frac{\ln 2+1}{2}-\frac{\ln N+1}{N}+\frac{\ln N}{N^2}.\)

So a lower bound is

\(\displaystyle \sum_{r=1}^{N}\frac{\ln r}{r^2}\gt \frac{\ln 2+1}{2}-\frac{\ln N+1}{N}+\frac{\ln N}{N^2}.\)