6
$\begingroup$

We just had the root test in class:

$\sum_{n=1}^\infty a_n$ (in $\mathbb R)$ converges if $\lim\limits_{n\rightarrow\infty}\sup\sqrt[n]{|a_n|}<1$

Why is it important to take the $\lim\sup$ and not taking just $\lim$? Any examples?

I've considered some series but with none of them I had a problem of taking $\lim$ instead of $\lim\sup$.

$\endgroup$
2
  • 5
    $\begingroup$ The limsup always exists, unlike ordinary limits. So, in most practical applications you can just use regular $\lim$. Fortunately, even when the limit doesn't exist, you can still get information via the limsup. $\endgroup$ Dec 14, 2012 at 14:23
  • $\begingroup$ Can you please be more specific? What does "limsup does always exist" mean? $\endgroup$
    – xotix
    Nov 12, 2017 at 14:09

3 Answers 3

7
$\begingroup$

If $\lim$ exists, it is of course the same as $\limsup$. The formula with $\limsup$ is useful when $\lim$ doesn't exist. Example: With $a_n=\frac{1+(-1)^n}2$ the expression$\sqrt[n]{|a_n|}$ has no limit, but the $\limsup$ is $1$.

$\endgroup$
2
$\begingroup$

$\limsup$ "contains" $\lim$ in the sense that if the limit exists, it coincides with $\limsup$.

So the root test also works when $\lim \sqrt[n]{|a_n|}$ doesn't exist, for example: $a_{n} = 2^{-n}$ if $n$ is even, and zero otherwise.

$\endgroup$
2
$\begingroup$

The key point in the proof of this theorem is that you compare your series with a convergent geometric series. Hence, you must make sure that eventually $|a_n| < q <1$ for some $q$ for all $n$ larger than some $N_0$. If you know the $\lim \sup$ satisfies $\lim\limits_{n\rightarrow\infty}\sup\sqrt[n]{|a_n|}<1$ then you can easily conclude this using some small enough neighbourhood. For this it is irrelevant whether the $a_n$ converge or not, you just need this estimate.

$\endgroup$

You must log in to answer this question.