calculus - Why $sum_{k=1}^{infty}frac{ln(k)}{k}$ diverges?

So my solution manual says this diverges since it is a p-series and $p=1 \le 1$

I understand why that p-series converges, but I don't understand conceptually how they got there using the comparison test (as the instructions advise).

Again, here is my series:

$$\sum_{k=1}^{\infty}\frac{\ln(k)}{k}$$

The instructions used the following comparison:

$$\frac{\ln(k)}{k} \gt \frac{1}{k}$$

This is what confuses me about this problem and the solution. Why was the above comparison used? I understand the concept that if $a_n \lt b_n$ and $b_n$ converges then $a_n$ converges.

This makes sense to me because I can see a graph in my head of a larger series converging so the smaller similar series must also converge.

Is the above comparison checking to see if the smaller series diverges and if so the larger must also diverge?

Please help me to understand this.

Answer

It is also true that if $a_n > 0, \quad b_n \gt 0$, $\sum a_n$ diverges and $a_n < b_n$, then $\sum b_n$ also diverges.

Here, we are comparing

$$a_n = \dfrac{1}{n}\;\;\text{ and }\;\;b_n = \dfrac {\ln n}n$$ and we have:

• $a_n >0, b_n>0$




• $a_n$ diverges; (it's the harmonic series, which diverges;
  
  you should make sure you know this - and if you haven't already
  explored the proof of its divergence, do so!),




• $b_n > a_n$. So our series with the term $b_n$ diverges.