Definition of Cauchy's Convergence Test using $\lim$ rather than $\limsup$

Question

I am looking at The ''Penguin Dictionary of Mathematics'', editions 2 and 4, edited by David Nelson (1998 and 2008 respectively).

I see the entry for "Cauchy convergence test" and I see this:

Let $a_1 + a_2 + \cdots + a_n + \cdots$ be an infinite series of positive terms. If $\displaystyle \lim_{n \to \infty} (a_n)^{1/n} < 1$ then the series converges. If the limit exceeds $1$, the series diverges.

This is at odds with the usual definition, which uses $\limsup$ instead of $\lim$.

(It is also the case that the usual definition does not specify positive terms, but summates over $|a_n|$ instead of $a_n$ which is a trivial distinction, so I'm not bothered about that so much.)

I don't understand what the consequences are of using $\lim$ instead of $\limsup$ as my general understanding of this whole area is shaky.

My question is: why is it necessary to use $\limsup$ not $\lim$, and what "goes wrong" if you try to use $\lim$?

I have not laid hands on any material which explains exactly why $\limsup$ not $\lim$ is used.

Answer 1

The test as you state it is correct, it is just less general than the one using $\limsup$. In fact, in response to the OP's comment, strictly speaking, it less general in two ways:

General test: It $(a_n)$ is a sequence of complex numbers then the series $\sum a_n$ converges absolutely if $\limsup_n |a_n|^{1/n}<1$ and diverges if $\limsup_n |a_n|^{1/n}>1$.

The test as stated from the Dictionary is less general because

$i)$ it requires the all of the $a_k$ to be positive, and

$ii)$ it requires that $\lim_n a_n^{1/n}$ exists.

That first restriction is a relatively minor one, as I expect the Dictionary is implicitly saying, since if you are given a series $\sum_{k \geq 1} a_k$ with complex terms $a_k$, then if $a_k$ is nonzero for only finitely many $k$, it obviously converges, while if not, we may set $b_n$ to be the modulus of the $n$-th nonzero term of the sequence $(a_n)$, so that $\sum a_n$ converges absolutely if and only if $\sum b_n$ does, and $\sum b_n$ has only positive terms.

The second difference however, is more significant, in that $\limsup_n |a_n|^{1/n}$ always exists (though it may be $+\infty$) whereas requiring $\lim |a_n|^{1/n}$ to exist is a genuine restriction.

To give an example, take the series $\sum_{n=0}^\infty a_n$ where $a_n=c^{n^2}$ unless $n=k!$ for some $k$, and in that case $a_{k!} = c^{k!}$ for some $c \in(0,1)$. Now if we take the function $x^{1/k}$ to be the inverse of the function $p_k\colon [0,\infty)\to [0,\infty)$ where $p_k(t)=t^k$, then $$ a_n^{1/n} =\left\{ \begin{array}{cc} c,& \exists k\in \mathbb N, n=k!\\ c^n, & \text{otherwise}. \end{array}\right. $$ Thus if $n_k=k!$ the $a_{n_k}^{1/n_k} = c$ for all $k$, so that $c$ is a limit of a subsequence of $(a_n)$. On the other hand, if $(p_k)$ denotes any increasing sequence of positive integers in $\mathbb N\backslash\{k!: k\in \mathbb Z_{\geq 0}\}$, e.g., the sequence of primes $(2,3,5,...)$, then $a_{p_k}^{1/p_k} =c^{p_k}\to 0$ as $k\to \infty$, so that $0$ is also a limit of a subsequence of $(a_n^{1/n})$ and hence the sequence $(a_n^{1/n})$ does not converge, and so the test of the Dictionary does not apply.

But now for all $n\in \mathbb N$ we have $a_n^{1/n} \in \{c,c^n\}$ and $\text{max}\{c,c^n\} = c$ as $c\in (0,1), n \geq 1$, hence it follows $a_n^{1/n}\leq c$ for all $n$, and thus $\limsup (a_n^{1/n})\leq c$. Since $c$ is also the limit of a subsequence of $(a_n^{1/n})$ it follows $\limsup (a_n^{1/n})=c$. Thus as $0<c<1$ the general form of the test shows the series converges.