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:


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.

  • 2
    $\begingroup$ "Is the above comparison checking to see if the smaller series diverges and if so the larger must also diverge?" Pre-cise-ly. $\endgroup$ – Daniel Fischer Nov 24 '13 at 20:07

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.

  • $\begingroup$ I don't understand how that is true. If $b_n$ is bigger than $a_n$ how can we know for sure that $b_n$ diverges? $\endgroup$ – hax0r_n_code Nov 24 '13 at 20:09
  • $\begingroup$ $a_n$ is the series that you chose to compare with, and thus you would already know if it converges/diverges. $\endgroup$ – Zhoe Nov 24 '13 at 20:13
  • 1
    $\begingroup$ Just to be picky, $b_n > a_n$ is only true for $n\geq 3$. $\endgroup$ – user940 Nov 24 '13 at 20:20
  • $\begingroup$ Duly noted, @Byron! $\endgroup$ – Namaste Nov 24 '13 at 20:21

Recall that a series $\displaystyle\sum_{n\ge0} a_n$ is said to be convergent if the partial sum sequence $\displaystyle\left(\sum_{k=0}^n a_k\right)_n$ is convergent. Now use what you know about the sequences and the fact that the harmonic series is divergent.


The comparison test that you would use for the convergence applies exactly the same when your series diverges, as seen on this article :


So your reasoning is perfectly correct ;) !


For another approach to this problem you could use the Integral Test. Consider $$\int_1^\infty{\ln(x)\over x} dx.$$ If $u=\ln(x)$, then $du={1\over x} dx$. Which gives us $$\lim_{t\rightarrow \infty}\int_1^t{\ln(x)\over x}dx=\lim_{t\rightarrow \infty}{\ln(x)^2\over 2}|_1^t=\lim_{t \rightarrow \infty}{\ln(t)^2\over 2}=\infty.$$ Thus $$\sum_{k=1}^\infty{\ln(k)\over k}$$ is divergent. If the divergence of the Harmonic series is causing an issue check out this link, hopefully it can help How do I prove that the series $\sum_{n=1}^\infty \frac{1}{n}$ diverges?.


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