8.2 Nonnegative Series 465~n. Then lim an = lim ( y'5rL - 10. yn) = lim ( v'5 n - lOyn) =
y ' " n->oo bn n--+oo 3n + yn 1 n--+oo 3n + yn}~~. (v'5-10/yn) v'5 v'5
3
+ l/yn = 3· Thus, p =
3
, and so by the limit comparisontest, the series (a) diverges.(b) In Exercise 6.4.19 we proved that ln n < n, so for large n the terms of
1 1
the series (b) are somewhere between 2 and -. Thus, it might be a good idea
n n
ln n. 1. ln n 1
to compare - 2 with 312. Lettmg an = - 2 and bn = 312 , we have
n n n n
lim an = lim (lnn. n3/ 2 ) = lim (lnn ) ·
n->oo bn n->oo n^2 1 n->oo nl/2
By L'Hopital's rule,
lim ( ln x ) = lim (l) = lim (
2
X->00 xl/2 X->00 _l_ X-> 00 VxX ) = X->00 lim (~) fx = O •
2y'X y.i,
1
In this case p = 0 and L n 312 is a convergent p-series (p = ~). Thus, Part(^00) ln n
(b) of the limit comparison test tells us that L - 2 converges. D
n=l n
THE RATIO TEST
Theorem 8.2.10 (Ratio Test, Basic Form) Let Lan be a series with
strictly positive terms.
(a) If ::3 0 < r < 1 such that an+l :::::; r for all but finitely many n, then Lan
an
converges.
(b) If an+l ~ 1 for all but finitely many n, then Lan diverges.
an
Proof. Let L an be a series with strictly positive terms.
(a) Suppose ::3 0 < r < 1 such that an+l :::::; r for all but finitely many n.
an
That. 1s, ::3 no EN 3 n ~no==> an--+l :::::; r. Then
an
ano+l :::::; ran 0
an 0 +2 :::::; rano+l :::::; r^2 an 0