Proof: Since
converges, is a null sequence, and hence
is bounded. Say
for all
. Let , so
, and let
. Then for all

Now is a convergent geometric series, so by the comparison test, converges; i.e., is absolutely convergent.

Proof: Suppose . If converges, then by the theorem, would also converge, contrary to our assumption.

- a) converges only when .
- b) converges for all .
- c) There is a number such that converges absolutely for and diverges for .

Proof: Suppose that neither a) nor b) is true. Then there are numbers such that converges and diverges. If , and , it follows that converges and diverges. By a familiar procedure, build a binary search sequence such that , and for all , converges and diverges. Let be the number such that . Then for all and . If , then for some we have , and

If , then for some we have , and

If a power series has radius of convergence
, I call the *disc of convergence* for the series, and I call the *circle of
convergence* for the series. If , I call
the disc of convergence of
the series (even though
is not a disc).

Hence

By the ratio test, is absolutely convergent if , and is divergent if . It follows that the radius of convergence for our series is .

- a)
- .
- b)
- .

- a)
- Find a power series whose radius of convergence is equal to .
- b)
- Find a power series whose radius of convergence is .
- c)
- Find a power series whose radius of convergence is .