This exercise sheet contains some additional questions that may be useful for revision or review of the material in these web pages.
This is work in progress. Please return here later.
Using the theorem on subsequences, and/or the theorem on boundedness of convergent sequences, and/or the theorem bounding the limit of a bounded sequence, prove the following sequences with th term as given do not converge.
(a) if is prime, otherwise.
(d) . (Hint: consider the subsequences and where and .)
Prove that for sufficiently large . (Take logs of both sides. What properties of the function do you require?)
Hence show that the series converges.
A function is defined by setting .
Prove that is defined for all (i.e., the limit above exists for all ) and takes the following values: if ; and if .
(Hint: Split into cases and divide top and bottom of the fraction by appropriate numbers. You may find it helpful to use that fact that if .)
Solution to Exercise 2, in the special case when : Note that for these values of we have as by standard null sequences , and continuity of the arithmetic operations. Thus for .
The next two exercises concern some common misconceptions and
errors in beginning analysis. Many of these examples are worth learning.
Some are easy but some are rather subtle and tricky. (One or two are
dreadful howlers that I hope no-one reading this would ever consider
writing.) If you have a good
feel for things like this you are
well on the way to being rather good at analysis.
The following statements are all false. In each case, find a counterexample.
(a) If the sequence diverges and the sequence diverges then their sum diverges.
(b) If the sequence diverges and the sequence diverges then their product diverges.
(c) If the sequence is a null sequence then converges.
(d) If for all and as then .
(e) If is a bounded sequence and has a convergent subsequence as then .
(f) If and both converge, then converges and equals .
(g) If is a null sequence then as .
(h) If the sequence does not converge then as .
(i) If and are both sequences and as then either or as .
(j) If as and converges, then converges. (Hint: consider an appropriate conditionally convergent series.)
(k) If as and converges, then converges. (Hint: let and .)
Solution to Exercise 3(a): Let the sequence be defined by and be defined by . Then both these sequences diverge, and yet their sum is or which defines the constant sequence with value 0, which converges.
Solution to 1(a): Let be the subsequence of consisting of all those with prime, and let be the subsequence of consisting of all those with non-prime. Both and are valid subsequences as there are both infinitely many primes in and infinitely many non-primes in . But is the constant sequence with value and is the constant sequence with value .
Therefore does not converge, for if as then and by the theorem on subsequences, so by the theorem on uniqueness of limits as since constant sequences converge to their value, and by the theorem on uniqueness of limits as as this is a constant sequence also. But this means , which is false.