# If $\limsup \frac{a_{n+1}}{a_n}<1$ does $\sum a_n$ converge even if $\lim\frac{a_{n+1}}{a_n}$ does not exist?

I was wondering why the Ratio test has the $$\lim$$ sign and the root test the $$\limsup$$ sign.

• Quotient test: $$\lim|\frac{a_{n+1}}{a_n}|<1\Rightarrow \sum a_n$$ converges.
• Ratio test: $$\limsup \sqrt[n]{|a_n|}<1\Rightarrow \sum a_n$$ converges.

If I pick $$a_n=\begin{cases}2^{-\frac{n}{2}} &\mbox{n even}\\3^{-\frac{n+1}{2}} &\mbox{n odd} \end{cases},$$ then for $$\frac{a_n+1}{a_n}$$ we either have $$(3/2)^{\frac{n+1}{2}}$$ or $$(2/3)^{\frac{n}{2}}\cdot 3$$. It is not bounded therefore the limit does not exist, and $$\limsup=\infty$$ makes no difference.

The textbook says that this example shows that a Quotient test analogously to the Ratio test with $$\limsup|\frac{a_{n+1}}{a_n}|$$ instead of $$\lim|\frac{a_{n+1}}{a_n}|$$ is not true.

What does this sentence mean why does it justify that if we have $$\limsup<1$$ we do not have necessarily $$\sum a_n<\infty$$. Because that is what I am getting out of it.

Edit:

I have understood it now the Ratio test also says if $$\lim \frac{a_{n+1}}{a_n}>1$$ the series diverges. If I would change this with $$\limsup$$ then we would get a contradiction with the root test

• Yes. Look carefully at the proof of the ratio test and you'll see the limit is not required to exist. Mar 22, 2019 at 15:49
• @new2math Please let me know how I can improve my answer. I really want to give you the best answer I can. Apr 3, 2019 at 3:31
• Does this answer your question? Ratio test with limsup vs lim Mar 22, 2023 at 9:56

The ratio test can be expressed as follows.

Let $$\ell=\liminf_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$$ and let $$L=\limsup_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$$. Then, the series $$\sum_{n=1}^\infty a_n\begin{cases} \text{converges (absolutey)}&, L<1\\\\ \text{diverges }&, \ell>1\\\\ \text{diverges }&, \left|\frac{a_{n+1}}{a_n}\right|\ge1\,\text{for all large}\,n\\\\ \text{inconclusive}&,\text{otherwise} \end{cases}$$

The root test is stronger than the ratio test since

$$\liminf_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\le \liminf_{n\to \infty}\sqrt[n]{|a_n|}\le \limsup_{n\to\infty}\sqrt[n]{|a_n|}\le \limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$

In the example in the OP, the ratio test in inconclusive since $$L=\infty$$ and $$\ell=0$$. However, the root test reveals

$$\limsup_{n\to\infty}\sqrt[n]{|a_n|}=2^{-1/2}<1$$

and the series converges.

If $$\limsup_n\left\lvert\frac{a_{n+1}}{a_n}\right\rvert<1$$, then take some $$c\in\left(\limsup_n\left\lvert\frac{a_{n+1}}{a_n}\right\rvert,1\right)$$. Then $$\limsup_n\left\lvert\frac{a_{n+1}}{a_n}\right\rvert and so, for some $$N\in\mathbb N$$, if $$n\geqslant N$$, then $$\left\lvert\frac{a_{n+1}}{a_n}\right\rvert. But then $$\lvert a_{N+1}\rvert, $$\lvert a_{N+2}\rvert and so on. So, yes, the series $$\sum_{n=0}^\infty a_n$$ converges absolutely. There is no difference between the ratio test and the quotient test as far as this is concerned.