Previous page (Exercises 5) | Contents | Next page (Exercises 7) |

- For each of the following sequences, decide whether they are eventually monotonic (and prove it).

(a) ( (*n*+1)/(*n*+2) )

(b) (*n*+ 8/*n*)

(c) (*n*+ (-1)^{n})

(d) (2*n*+ (-1)^{n}) - Let
*s*_{n}be the partial sum of the series (1/*i*^{2}). Prove that the sequence (*s*_{n}) is monotonic and use induction to show that*s*_{n}≤ 2 -^{1}/_{n}. Hence prove that the series converges.

Leonhard Euler (1707 to 1783) showed that its limit is π^{2}/_{6}. - Define a sequence (
*a*_{n}) by*a*_{1}= 1 and*a*_{n+1}=*a*_{n}+^{1}/_{n}for*n*≥ 1. Prove that the limit as*n*→ ∞ of |*a*_{n}-*a*_{n+1}| is 0 but that (*a*_{n}) is*not*a Cauchy sequence. - Let (
*s*_{n}) be the sequence of partial sums of the*positive*series*b*_{n}. If 0 ≤*b*_{n}≤*c*_{n}and*c*_{n}has a convergent sequence of partial sums, deduce that (*s*_{n}) is monotonic and bounded above and hence convergent.

This is the so called**Comparison Test for positive series**discovered by Jacob Bernoulli (1654 to 1705). - Let 0 <
*r*< 1 and let (*a*_{n}) be a sequence for which |*a*_{n}-*a*_{n+1}| <*r*^{n}for all*n*. Prove that (*a*_{n}) is a Cauchy sequence.

Define a sequence (*a*_{n}) by*a*_{1}= 0,*a*_{2}= 1 and*a*_{n+1}=*a*_{n}/3 + 2*a*_{n-1}/3 for*n*> 1. Use the last result to show that this is a Cauchy sequence and hence prove that it is convergent.

Any guesses about what it converges to? - For any
*x*> 0 we have*x*^{0}= 1 and 0^{x}= 0. So is the limit as*x*→ 0+ of*x*^{x}= 0 or 1 or neither?

What is the limit as*n*→ ∞ of*n*^{1/n}? - If you invest £
*N*at a rate of (say)12% per annum*simple interest*, then at the end of the year you would have £*N*(1 + 0.12).

If the interest were calculated twice (i.e. the rate is 6% per 6 months) you would have £*N*(1 + 0.06)^{2}at the end of the year. (This is called*compound interest*.)

If the interest were calculated 3 times a year, you would have £*N*(1 + 0.04)^{3}at the end. And so on.

Compounding more and more often, the number of pounds you would have at the end of the year would be the limit as*n*→ ∞ of the sequence (*a*_{n}) with*a*_{n}=*N*(1+0.12/*n*)^{n}.

Take logarithms, replace 1/by*n**x*and use l'Hôpital's rule to find the limit as*x*→ 0.

Previous page (Exercises 5) | Contents | Next page (Exercises 7) |