sequences and series - Comparison test $sum_{k=1}^{infty} frac{1 + ln(k)}{k}$

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

Determine whether it converges or diverges.

I don't think I could do limit comparison test because the $\ln(k)$ messed me up. Pretty sure I could do this with integral test but I think this is possible with comparison test as well. Could someone tell me if it is? For instance I'm looking for a $b_k$ value that is

$$0 \leq a_k \leq b_k$$

My textbook uses $b_k = \frac{1}{k}$, but how is the hypothesis met with this?

$$\frac{1+\ln(k)}{k} \,\geq\, \frac{1}{k}$$

Thats wrong it should be $a_k \leq b_k$ $\forall n \geq 1$

Answer

The book is correct. Note that $1+\log(k)\ge 1$ for all $k\ge 1$. Hence,

$$\frac{1+\log(k)}{k}\ge \frac1k$$

Since the harmonic series diverges, then the series $\sum_{k=1}^\infty \frac{1+\log(k)}{k}$ diverges by comparison. That is to say, the series of interest dominates the divergence harmonic series.