One of the examples in my textbook uses the comparison
$$\frac{1}{\sqrt{k} \ln{k}} \geq \frac{1}{k}$$
This comparison is not immediately obvious to me. I can understand that $\frac{1}{\sqrt{k}} \geq \frac{1}{k}$ holds but I would have assumed that by multiplying the denominator of the RHS by $\ln{k}$ the inequality would be reversed.
Let $F (x)=2\ln (x)-x $ for $x>2 .$
$$F'(x)=\frac {2-x}{x}<0$$ thus $F $ is strictly decreasing and
$$(\forall x>2)\;\;\;F (x)<F (2)<0$$ then, for $k>4$ $$F (\sqrt{k})<0 \implies 2\ln (\sqrt {k})<\sqrt {k }$$
$$ \implies \ln (k)<\sqrt {k}$$
$$ \implies \sqrt {k}\ln(k)<k $$ $$\implies \frac {1}{\sqrt {k}\ln (k)}>\frac {1}{k} $$ the positive series $\sum \frac {1}{\sqrt {k}\ln (k)} $ is divergent.
This is equivalent to the statement that $$\sqrt k \ln k \leq k$$ which immediately follows from the fact that $\ln k \leq \sqrt k$. To show that this holds for $k\geq 4$, define $f(k) = \sqrt k - \ln k$. Then $f(4) > 0$, and $$f'(k) = \frac{1}{2\sqrt k} - \frac{1}{k} \geq 0$$ So, $f$ is nondecreasing for $k \geq 4$, and so $\ln k \leq \sqrt k$. This is sufficient to apply the comparison test.
Theorem. If $a > 0$, then $\log (x) < x^{a}$ for large $x$.
It follows that $\sqrt{k}\log k \leq \sqrt{k}\sqrt{k} = k$ for large $k$; then the inequality under consideration follows.
