SECTION 5.2 Numerical Series 275
nlim→∞
an+1
an
= limn→∞
Å(n+2) 3
(n+1)!
ã
Å(n+1) 3
n!
ã
= limn→∞
(n+ 2)^3
(n+ 1)(n+ 1)^3
= 0.
Therefore,
∑∞
n=1
(n+ 1)^3
n!
converges.
Example 8. Consider the series
∑∞
n=1
n^2
3 n
. We have
nlim→∞
an+1
an
= limn→∞
Å(n+1) 2
3 n+1
ã
(n 2
3 n
)
= limn→∞
(n+ 1)^2
3 n^2
=
1
3
< 1.
It follows, therefore, that
∑∞
n=1
n^2
3 n
also converges.
Example 9. Consider the series
∑∞
n=1
n!
2 n
. We have
nlim→∞
an+1
an
= limn→∞
Å(n+1)!
2 n+1
ã
(n!
2 n
)
= limn→∞
n+ 1
2
=∞,
which certainly proves divergence.
Exercises
- Test each of the series below for convergence.
(a)
∑∞
n=1
n+ 2
n^2 + 10n (b)
∑∞
n=0
n^2 −n+ 2
n^4 +n^2 + 1