14.3. SERIES AND SEQUENCES OF LINEAR OPERATORS 397

It follows |y(t)|2 = e2at , |y(t)|= eatas claimed.This follows because in general, if

z′ = cz, z(0) = 1,

with c real it follows z(t) = ect . To see this, z′− cz = 0 and so, multiplying both sides bye−ct you get

ddt

(ze−ct)= 0

and so ze−ct equals a constant which must be 1 because of the initial condition z(0) = 1. ■

Definition 14.3.6 The function in Corollary 14.3.5 given by that power series is denotedas

exp(λ t) or eλ t .

The next lemma is normally discussed in advanced calculus courses but is proved herefor the convenience of the reader. It is known as the root test.

Definition 14.3.7 For {an} any sequence of real numbers

lim supn→∞

an ≡ limn→∞

(sup{ak : k ≥ n})

Similarlylim inf

n→∞an ≡ lim

n→∞(inf{ak : k ≥ n})

In case Anis an increasing (decreasing) sequence which is unbounded above (below) thenit is understood that limn→∞ An = ∞ (−∞) respectively. Thus either of limsup or liminfcan equal +∞ or −∞. However, the important thing about these is that unlike the limit,these always exist.

It is convenient to think of these as the largest point which is the limit of some sub-sequence of {an} and the smallest point which is the limit of some subsequence of {an}respectively. Thus limn→∞ an exists and equals some point of [−∞,∞] if and only if the twoare equal.

Lemma 14.3.8 Let{

ap}

be a sequence of nonnegative terms and let

r = lim supp→∞

a1/pp .

Then if r < 1, it follows the series, ∑∞k=1 ak converges and if r > 1, then ap fails to converge

to 0 so the series diverges. If A is an n×n matrix and

r = lim supp→∞

||Ap||1/p , (14.7)

then if r > 1, then ∑∞k=0 Ak fails to converge and if r < 1 then the series converges. Note that

the series converges when the spectral radius is less than one and diverges if the spectralradius is larger than one. In fact, limsupp→∞ ||Ap||1/p = limp→∞ ||Ap||1/p from Theorem14.2.4.