How to prove that $\lim_{k\rightarrow\infty}\ln k - \frac{1}{2}H(k)$ diverges?

Question

It is known that $$\lim_{k\rightarrow\infty}\ln k - H(k)=\gamma$$(Where $H(k)$ is the $k$th harmonic number and $\gamma$ is Euler's constant). I decided to put a random $\frac{1}{2}$ in the front of $H(k)$ and I got some limit that seems to diverge when graphed. So how do I prove that $$\lim_{k\rightarrow\infty}\ln k - \frac{1}{2}H(k)$$ diverges? I was able to show that it is greater than $-\gamma$ which is a bit obvious, but that's it.

Answer 1

$$ \ln k - \frac{1}{2}H(k) = (\ln k - H(k)) + \frac{1}{2}H(k) $$ is divergent as the sum of a converging and a diverging sequence.