Why L < 1 forces convergence
As with the ratio test, the trick is to trap the tail under a convergent geometric series —
only here a single root, not a chain of ratios, does the work. Suppose
L = \lim_{n\to\infty} |a_n|^{1/n} < 1.
Step 1 — choose r between L and
1.
L < r < 1.
Step 2 — the roots eventually drop below r. Since
|a_n|^{1/n} \to L < r, past some index
N the roots never exceed r:
|a_n|^{1/n} \le r \qquad \text{for all } n \ge N.
Step 3 — raise both sides to the nth power. This
is the move the root test is built for — the exponent comes right off:
|a_n| \le r^{\,n} \qquad \text{for all } n \ge N.
Step 4 — dominate the tail by a geometric series. Sum from
N onward and compare term by term to a geometric series of ratio
r < 1:
\sum_{n=N}^{\infty} |a_n| \;\le\; \sum_{n=N}^{\infty} r^{\,n} = \frac{r^{N}}{1 - r} < \infty.
Step 5 — conclude absolute convergence. The tail is finite, and the finitely
many earlier terms add only a finite amount, so
\sum_{n=1}^{\infty} |a_n| < \infty.
If L > 1, then |a_n|^{1/n} > 1 infinitely
often, so |a_n| > 1 infinitely often — the terms do not tend to
zero, and the
divergence test
finishes the job.
For a series \sum a_n, let
L = \displaystyle\lim_{n\to\infty} |a_n|^{1/n} (use
\limsup if the plain limit does not exist). Then:
-
L < 1 \Rightarrow
\sum a_n converges absolutely (tail dominated
by \sum r^n, r \in (L,1)).
-
L > 1 or L = \infty
\Rightarrow \sum a_n
diverges (terms do not tend to zero).
-
L = 1 \Rightarrow
inconclusive (e.g. every p-series gives
L = 1).
The two tests usually agree, and when both ratios and roots converge they give the
same L — there is a theorem that
\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = L \;\Longrightarrow\; \lim_{n\to\infty} |a_n|^{1/n} = L,
but not conversely. The root test can succeed where the ratio test stalls. The
classic witness has terms that alternate two recipes,
a_n = \begin{cases} 2^{-n}, & n \text{ even},\\[2pt] 3^{-n}, & n \text{ odd}. \end{cases}
The ratios a_{n+1}/a_n swing wildly between huge and
tiny and never settle, so the ratio test has no limit to read. But the roots are
tame — |a_n|^{1/n} is either \tfrac12
or \tfrac13, so \limsup = \tfrac12 < 1
and convergence is immediate. Whenever the term is naturally an
nth power, the root test is the sharper instrument — reach for it
on terms like \big(\tfrac{n}{2n+1}\big)^{n}, where one root
collapses the whole thing.