Advanced High-School Mathematics

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

1. 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