# Is it true? If $\lim_{n\to\infty}\frac{|a_{n}|}{|c_{n}|}=0$, then the series $\sum_{n=1}^\infty a_n$ converges

Let $$\{c_n\}$$ be a sequence that converges to $$0$$ and let $$\{a_n\}$$ be a sequence for which the following applies: $$\lim_{n\to\infty}\frac{|a_{n}|}{|c_{n}|}=0$$. Then the series $$\sum_{n=1}^\infty a_n$$ converges.

I tried to think of a counter-example to show that this is false but I can't really come up with any. I think it's false, because the only thing that this statement tells me is that the sequence $$\{a_n\}$$ converges faster to $$0$$ than $$\{c_n\}$$. And this doesn't really tell me anything about the convergence of the series.

Any help would be appreciated, thanks!

• You're right, it's false! $c_n=1/\sqrt{n}$ and $a_n=1/n$ will do. $a_n/c_n=1/\sqrt{n}$ which goes to $0$ as $n\to\infty$ but $\sum a_n$ is the harmonic series.
– EBO
Commented Feb 24, 2020 at 22:16
• Apparently, you can tweak the statement to make it true. Have a look at this. It's important to see why our example does not work, though; both of our sequences define non-convergent series. Namely our series defined by the $c_n$. Remember that $\sum 1/n^\alpha$ converges for $\alpha>1$ only.
– EBO
Commented Feb 24, 2020 at 22:30

Note that in case $$\sum c_n$$ is absolutely convergent, we can conclude with the argument below:
$$\lim_{n\to\infty}\frac{|a_{n}|}{|c_{n}|}=0$$ means that for $$n_0$$ large enough we have $$\forall n\ge n_0, \dfrac{|a_n|}{|c_n|}<1$$.
In particular $$\sum\limits_{n_0}^{\infty}|a_n|<\sum\limits_{n_0}^{\infty}|c_n|<+\infty$$ thus $$\sum a_n$$ is absolutely convergent too.
In case of semi-convergence of the series you can take $$c_n=\dfrac{(-1)^n}{\sqrt{n}}$$ and $$a_n=\dfrac 1n$$ as a counterexample.
And of course if you do not even suppose that $$\sum c_n$$ converge, there is really no reason it should work in general.
For instance take series like $$\sum\frac 1n$$, $$\sum\frac 1{n\ln(n)}$$, $$\sum \frac 1{n\ln(n)\ln(\ln(n))}$$, $$\cdots$$ none is convergent despite terms being little-o of the previous one.