# Why does Cauchy's Root Test for convergence of infinite series require $\limsup$?

I'm confused about the reasoning behind Cauchy's root test for convergence of infinite series. It states that for any series $\{a_n\}$, if $C = \limsup_{n\rightarrow\infty}{\sqrt[n]{|a_n|}} < 1$, then the series is absolutely convergent. If $C > 1$ the series diverges, and otherwise the test is inconclusive.

The proof states that if $\sqrt[n]{|a_n|} < k < 1$, $\forall_n \ | \ n > N$, then it is also true that $|a_n| < k^n$. This implies that $\sum_{i=N}^\infty{|a_i|}$ and $\sum_{i=N}^\infty{k^i}$ either both converge or diverge, and since $k < 1$, the second series converges and thus so does $\sum{|a_n|}$ (since $\sum_{i=0}^N{|a_i|}$ will always be finite).

This proof seems to make sense to me for $C = \lim_{n\rightarrow\infty}{\sqrt[n]{|a_n|}}$, why does it require $\limsup$?

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$\limsup$ always exists, $\lim$ does not always exist. – Daniel Fischer Aug 29 '14 at 23:51
@DanielFischer I actually know very little about $\limsup$, could you please elaborate on why? – user3002473 Aug 29 '14 at 23:52
$\limsup$ is a limit of a decreasing sequence. Thus by the monotone convergence theorem it has a limit (possibly an infinite one) for any sequence. – Ian Aug 29 '14 at 23:59
$\limsup x_n$ is the largest accumulation point of the sequence $(x_n)$ [possibly $+\infty$]. Every sequence of real numbers has an accumulation point in $[-\infty,+\infty]$, and it converges [improperly, possibly] if and only if it has only one accumulation point. If the sequence converges, then $\limsup x_n = \lim x_n = \liminf x_n$. Otherwise, we have $\liminf x_n < \limsup x_n$, and for every $k > \limsup x_n$, there is an $N$ such that $x_n < k$ for all $n\geqslant N$ (analogously for $c < \liminf x_n$, then $x_n > c$ for all large enough $n$). – Daniel Fischer Aug 29 '14 at 23:59
That makes sense, thank you both! – user3002473 Aug 30 '14 at 0:08