# 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 $a_k \leq b_k$ $\forall n \geq 1$

• The harmonic series diverges and this each terms of this series is greater than that of the harmonic series. – Alex Vong Apr 13 '17 at 2:44
• If you insist of using the inequality $0\leq a_k\leq b_k$, then you should be looking for $a_k$ and not $b_k$. This is the trick if you prove for divergence. So, we can take $a_k=\frac{1}{k}$ and $b_k=\frac{1+\ln k}{k}$ and then apply comparison test for divergence. – Juniven Apr 13 '17 at 2:45
• There is a divergence comparison test? – user349557 Apr 13 '17 at 2:51
• @user29418 Not for k=1. – TMM Apr 13 '17 at 2:51
• The only definition for comparison test I know is: Assuming all these are series If $0 \leq a_n \leq b_n, \forall n \in \mathbb N$, then $\sum b_n$ converges. Same with diverges. – user349557 Apr 13 '17 at 2:52

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.
• Hi Dr. MV. I thought it says by the definition that you have to find a value $b_k$ that is greater than or equal to $a_k$ and both are positive, then we can use $b_k$ to find the result? So im wondering why this is allowed. For example $$\sum_{n=1}^{\infty} \frac{tan^{-1}(n)}{n^{3/2} + sin^2 (n)} \leq \sum_{n=1}^{\infty} \frac{\pi/2}{n^{3/2}}$$. The textbook does it the exact same way you do it but I don't understand why this is allowed? – user349557 Apr 13 '17 at 2:48
• For $0\le a_k \le b_k$, we can show that the series $\sum a_k$ converges by showing that $\sum b_k$ converges. And the converse is also true. That is what we used here. We have $a_k\ge b_k\ge0$ with $\sum b_k$ divergent. – Mark Viola Apr 13 '17 at 2:54