2
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ The harmonic series diverges and this each terms of this series is greater than that of the harmonic series. $\endgroup$ – Alex Vong Apr 13 '17 at 2:44
  • $\begingroup$ 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. $\endgroup$ – Juniven Apr 13 '17 at 2:45
  • $\begingroup$ There is a divergence comparison test? $\endgroup$ – user349557 Apr 13 '17 at 2:51
  • $\begingroup$ @user29418 Not for k=1. $\endgroup$ – TMM Apr 13 '17 at 2:51
  • $\begingroup$ 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. $\endgroup$ – user349557 Apr 13 '17 at 2:52
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user349557 Apr 13 '17 at 2:48
  • $\begingroup$ 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. $\endgroup$ – Mark Viola Apr 13 '17 at 2:54
  • $\begingroup$ I got the definition mixed up. Thanks for clearing it up $\endgroup$ – user349557 Apr 13 '17 at 2:56
  • $\begingroup$ You're welcome. My pleasure. -Mark $\endgroup$ – Mark Viola Apr 13 '17 at 3:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.