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Let's assume we are given a series $\sum\limits_{k=1}^{\infty}a_k$ which we want to check for convergence. Two scenarios:

1.) After some manipulations of $a_k$ we get something like: $a_k= \frac{1}{k}+f(k)=\frac{1}{k}+O\left(\frac{1}{k^2}\right)$. Now, if someone asks you to prove the divergence via comparison test then I would simply conclude (for a $k$ that is large enough):

$$f(k)= O\left(\frac{1}{k^2}\right)\implies |f(k)|\leq C\frac{1}{k^2}\implies -C\frac{1}{k^2}\leq f(k)\leq C\frac{1}{k^2}\\\implies \Big| \frac{1}{k}-C\frac{1}{k^2}\Big|= \frac{1}{k}-C\frac{1}{k^2}\leq\frac{1}{k}+f(k)$$ where $C>0$. Further, we know that $\sum\limits_{k=1}^{\infty} \frac{1}{k}-C\frac{1}{k^2}$ diverges and hence $\sum\limits_{k=1}^{\infty}a_k$ diverges due to comparison test.

Is this right?

2.) After some manipulations of $a_k$ we get something like: $a_k= g(k)=O\left(\frac{1}{k}\right)$. Now, if someone asks you to prove the divergence via comparison test then I would argue that for a $k$ that is large enough we run into trouble, namely:

$$g(k)= O\left(\frac{1}{k}\right)\implies |g(k)|\leq C\frac{1}{k}\implies -C\frac{1}{k}\leq g(k)\leq C\frac{1}{k}\\\implies \color{red}{\Big|?\Big|}\leq g(k)\leq C\frac{1}{k}$$ where $C>0$. So in this case we cannot construct another series which is divergent and could be used to perform the comparison test, right?

Although our series $\sum\limits_{k=1}^{\infty}a_k$ has an $a_k$ which has the same behavior like $\frac{1}{k}$ it doesn't help us to decide if the series is divergent or not, right?

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  • $\begingroup$ The comparison test works for series with constant sign. $\endgroup$ Nov 20, 2020 at 22:03
  • $\begingroup$ $O(1/k)$ doesn't allow you to prove divergence. $O()$ is an upper bound: $1/k^2$ is $O(1/k)$ too, for instance. $\endgroup$
    – Clement C.
    Nov 20, 2020 at 22:11
  • $\begingroup$ @ClementC. so my ideas in the examples are correct? $\endgroup$
    – Philipp
    Nov 20, 2020 at 22:21

1 Answer 1

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With $$a_k=\frac{\sin(k)}{k}$$ and

$$b_k=\frac{|sin(k)|}{k}$$ we have

$$a_k=O(\frac 1k)\;\; and \;\; b_k=O(\frac 1k)$$ but $$\sum a_k \;\; converges$$ while $$\sum b_k \;\; diverges$$

What we know is, if $ C>0 $,

$$ a_k\le \frac{-C}{k}\implies \sum a_k\; \; diverges$$ and

$$\frac{C}{k}\le a_k\implies \sum a_k\;\; diverges$$

but

$$\frac{-C}{k}\le a_k\le \frac{C}{k}$$ does not imply the divergence.

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  • $\begingroup$ Yes this example is clear to me. However, I was trying to explain in a more general way why one can't use $O(\frac{1}{k})$ to conclude divergence. So I want to know if my reasoning is correct. $\endgroup$
    – Philipp
    Nov 20, 2020 at 22:30
  • $\begingroup$ @Philipp I just added some lines. i hope will help. $\endgroup$ Nov 20, 2020 at 22:39

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