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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

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

2 Answers 2

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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.

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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 $\limsup_n\left\lvert\frac{a_{n+1}}{a_n}\right\rvert<c$ and so, for some $N\in\mathbb N$, if $n\geqslant N$, then $\left\lvert\frac{a_{n+1}}{a_n}\right\rvert<c$. But then $\lvert a_{N+1}\rvert<c\lvert a_N\rvert$, $\lvert a_{N+2}\rvert<c^2\lvert a_N\rvert$ 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.

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