Let HN = $\sum_{n=1}^{\infty } \frac{1}{n} =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ . . .$ is a popular HP which is divergent infinite series (divergent means limit of this function won't be a finite number).
It's divergent nature can be prove by 'Comparison Test' , which proves that sum of elements of this infinite series would be infinite.
https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)
Another way to prove it's divergent nature is by 'Integral Test' , in which we'll prove that $\int_{1}^{\infty } \frac {1}{x}dx$ would be logn and we know when n $\rightarrow \infty then $ logn $\rightarrow \infty$
a and e ) Both function AN and BN would tend to $\infty$ when N$\rightarrow \infty$
c and d ) By using map we can compare both function HN and BN
We can easily see if we find area using $\sum$ the rectangular will leads to some extra area than the area which will be covered using $\int_{}^{}$ so we can say from here HN > BN
in our question here AN is discrete sum of our HP series , excluding 1 so its basically HN-1
so we have proved yet that logn=BN <HN=AN +1
Another way to prove lower bound and upper bound of HN is like this
As $y=\frac {1}{x}$ is a Monotonically Decreasing function
means if some X ∈ [K,K+1] then $\frac{1}{k+1} \leq \frac{1}{x} \leq \frac{1}{K}$
So $\sum_{K=1}^{n}\frac {1}{K+1} \leq \int_{1}^{n}\frac{1}{x} \leq \sum_{K=1}^{n} \frac {1}{K}$
So AN $\leq$ logn $\leq$ HN
so eventually logn $\leq$HN $\leq$logn+1
Useful reference
https://math.dartmouth.edu/archive/m105f13/public_html/m105f13notes1.pdf