The root test has to do with when a series of real or complex numbers converges. I am assuming the reader
has been exposed to infinite series. However, this that I am about to explain is a little more
general than what is usually seen in calculus. If you have a sequence of real numbers

{ak}

_{k=1}^{∞},
if

An ≡ supak
k≥n

then the sequence

{An}

is decreasing. In the above, sup_{k≥n}a_{k} means the least upper bound of all a_{k} for
k ≥ n or if there is no upper bound, A_{n} is simply said to equal ∞. This is just a formality to make it easy
to give an easy discussion. Then, since

{An}

is a decreasing sequence, there are two cases. One is that it is
bounded below and the other case is that it isn’t. In the first case, the sequence must converge to the
greatest lower bound of the A_{n} and in the second case, we say that the sequence converges to −∞.
Then

( )
lim sup a ≡ lim sup a
n→∞ n n→∞ k≥n k

Thus, if limsup_{n→∞}< r, it follows that for all n large enough every a_{k}< r. If limsup_{n→∞}a_{k}> r, it means
there are infinitely many k such that a_{k}> r.

Theorem 8.11.1Let a_{k}∈ F^{p}, F is either ℝ or ℂ and consider∑_{k=1}^{∞}a_{k}. Then this series convergesabsolutely if

1∕k
lim ksu→p∞ |ak| = r < 1.

The series diverges spectacularly if limsup_{k→∞}

|a |
k

^{1∕k}> 1 and iflimsup_{k→∞}

|a |
k

^{1∕k} = 1, the testfails.

Proof:Suppose first that limsup_{k→∞}

|ak|

^{1∕k} = r < 1. Then letting R ∈

(r,1)

, it follows from the
definition of limsup that for all k large enough,

|ak|1∕k ≤ R

Hence there exists N such that if k ≥ N, then

|ak|

≤ R^{k}. Let M_{k} =

|ak|

for k < N and let M_{k} = R^{k} for
k ≥ N. Then

∞∑ N∑−1 RN
Mk ≤ |ak|+ 1-− R < ∞
k=1 k=1

and so, by the Weierstrass M test applied to the series of constants, the series converges and also converges
absolutely. If

lim sup |ak|1∕k = r > 1,
k→∞

then letting r > R > 1, it follows that for infinitely many k,

k
|ak| > R

and so there is a subsequence which is unbounded. In particular, the series cannot converge and in fact
diverges spectacularly. In case that the limsup = 1, you can consider ∑_{n=1}^{∞}

1
n

which diverges by calculus
and ∑_{n=1}^{∞}

1
n2

which converges, also from calculus. However, the limsup equals 1 for both of these.
■

This is a major theorem because the limsup always exists. As an important application, here is a
corollary.

Corollary 8.11.2If∑_{k}a_{k}converges, then limsup_{k→∞}

|a |
k

^{1∕k}≤ 1.

If the sequence has values in X a complete normed linear space, there is no change in the
conclusion or proof of the above theorem. You just replace