Manipulating a series with big-O-notation

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?

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

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.

• 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. Nov 20, 2020 at 22:30
• @Philipp I just added some lines. i hope will help. Nov 20, 2020 at 22:39