Finite harmonic sum inequality I wish to prove some inequality involving a finite harmonic series: 
$$\sum_{k=n+1}^{n^2}\frac{1}{k}>\sum_{k=2}^{n}\frac{1}{k}$$
Certainly $\frac{1}{nk+q}≥\frac{1}{n(k+1)}$
for $q=1,2,3,....,n.$
So that $$\sum_{q=1}^n\frac{1}{nk+q}≥\frac{1}{k+1}$$
Adding the last inequality from $k=1$ to $n-1$  should yield the required inequality but I don't see how it does.

Let $H_n=\sum_{k=1}^n \frac1k$. Then we want to prove $H_{n^2}-H_n\geq H_n-1$, which is equivalent to $$H_{n^2}\geq 2H_n-1.$$
Indeed, using $\ln(n)+\gamma\le H_n\le \ln(n)+\gamma+\frac1{2n}$ (it is a weaker version of equation (8)), we get $$H_{n^2}\geq \ln(n^2)+\gamma=2\ln(n)+\gamma$$ and $$2H_n -1\le \frac1n+2\gamma +2\ln(n)-1$$ so it remains to prove that $$\frac1n+\gamma\le1,$$ which is indeed true for $n\geq 3$ (note that $\gamma\approx0.57$ is the Euler-Mascheroni constant). The cases $n=1,2$ can be checked manually.
