A sequence is given by
(a) Prove that
for all .
(b) Deduce that for all .
(c) Using induction, show that
for each .
(d) Using deduce that as .
Suppose in each case that is a sequence whose th term is given by the expressions below. Prove that converges and find the limit. If you use the continuity of any standard functions in your answer, state which function(s) you use that you know to be continuous and at what value or the continuity of that(those) function(s) is required.
(a) ; (b) ; (c) ; (d) .
Prove that the following sequences do not converge, using the following method.
Suppose first that the sequence in question converges to some . Then find convergent sequences , , ..., and continuous functions such that some expression involving evaluates to a sequence that you in fact know to be non-convergent.
The convergence or nonconvergence of any sequences you use must be justified either by stating that they are "standard" convergent/nonconvergent sequences discussed in lectures (such as or ) or by giving a proof.
The sequence is defined inductively by and . Prove the following by induction on .
(a) for all .
(b) for all .
(c) Prove the following holds for all : .
(d) Say whether has a limit in and if it does, find this limit.
The sequence is defined by , , and . Prove the following.
(a) for all .
(b) for all .
(c) The subsequences and are both monotonic. (Hint: use (b) above to find a formula for .)
(d) Explain why these results imply that and both converge, and in fact both converge to the same limit . Which theorem from the course allows you to deduce that is the limit of the original sequence ?
(e) Find the limit . Suggestion: Use (b) repeatedly to get an expression for in terms of . Sum this series then find the limit.
Solution to 3.2(a). Write
and observe that and as and so by the constant sequences and the continuity of , we have as . Also by the constant sequences and the continuity of again, we have as . By the continuity of division at we have
Solution to 3.3(a). Suppose defines a sequence converging to . Consider
Now is a constant sequence, so converges to . The sequence converges to by a result from lectures. So by the continuity of at as , by the continuity of at , as and by the continuity of division at , the sequence above converges to . But this is impossible as
which is known from lectures to be nonconvergent.
Solution to 3.4. (a) Let be the statement
using and , which say that and . So is true. This completes the proof by induction.
(Note: If you are not happy about assuming
an induction hypothesis that says that
are all true, think of this
argument just given as showing that holds and
where the statement is
(b) Let be the statement
by and , so is true. This completes the proof by induction.
(c) Let the statement be , so as , says which is true. Now assume is true for and compute:
as . Also,
as and . This proves . So is true for all , by induction.
(d) These results show that is bounded (by (a) and (b)) and is monotonic (by (c)), and so by monotone convergence has a limit . Now and are subsequences of so by the theorem on subsequences converge to the same limit . But , so by the continuity of + and division by 4, as . By uniqueness of limits we have , and solving this we get . Hence .