M1P1 Comments on Sheet 2 1. (a) We guess l = 0. Fix > 0. Here |an - l| = 1/ n, so we have |an - l| < n > 1/2 . Let N be any integer such that N > 1/2 . Then using (1), we find n N = n > 1/2 = |an - l| < , as required. (b) Similar; here l = 1, |an - l| = 1
M1P1 Comments on Sheet 3
1
1. Answers are (i) 3 (ii) 2 (iii) 2 . For (iii), note that the nth term in the sequence is equal
to 1 (1 + 1/n).
2
2. Possible examples are:
(a) an 1, bn = (1)n .
(b) an 0, bn = n.
(c) an = bn = (1)n .
3. Yes, because bn = (an +
M1P1 Comments on Sheet 4
1. Fix any > 0. Since an l, we can choose N such that n N = |an l| < . Since
for any n we have |an | |l| |an l|, it follows that n N = |an | |l| < . By
denition, this shows that |an | |l|.
2. |an | 0 means: given any > 0, there is
M1P1 Comments on Sheet 5 1. Fix > 0. Since (an ) and (bn ) are Cauchy, and /2 > 0, there are indices N1 and N2 such that n, m N1 = |an - am | < /2 and n, m N2 = |bn - bm | < /2 . Now, by the triangle inequality, |(an + bn ) - (am + bm )| |an - am | + |bn
M1P1 Comments on Sheet 6
1. Suppose that an 0. Since 1 > 0, the basic denition (1.1) tells us that there is an
N such that n N = |an | < 1; in particular, |aN | < 1. This is a contradiction, since
we were given that |an | 1 for all n; hence (an ) does not
M1P1 Comments on Sheet 7 1. (a) |un+1 /un | = 1 (1 + 1/n), which 2 (b) |un+1 /un | =
1 (1 3 2 1 2
< 1; by the ratio test, the series converges.
1 3
+ 1/n) , which
< 1; by the ratio test, the series converges.
(c) un 1 for all n, so the sequence (un ) doe
M1P1 Comments on Sheet 8
1. In each case, let un denote the nth term of the series.
(a) |un+1 /un | = |z |(1 + 1/n)3 |z |. So by the ratio test the series converges if |z | < 1
and diverges if |z | > 1; that is, the radius of convergence R is 1.
(b) |un+1
M1P1 Comments on Sheet 9
1. Possible examples are:
(a) the G.P.
zn;
1
(b) the series
1
(c) the series
1
z n /n2 (it is absolutely convergent for |z | = 1;
(1)n z n /n.
2. 1.
3.
0
(n + 1)z n . Our original series converges absolutely to (1 z )1 for |z | <
M1P1 Comments on Sheet 10
1. Given any > 0, let = 2 . Then > 0, and if 0 < |x 3| < , then |f (x) 5| =
|2x 6| = 2|x 3| < 2 = . By denition, this means that f (x) 5 as x 3. Notice
that the information given about f (3) is completely irrelevant.
2. Fix any >
M1P1 Comments on Sheet 1
1. Assume that |x| < for every > 0 and suppose that x = 0. Then |x| > 0, so we
get the contradiction |x| < |x|. Hence x = 0. It follows that the sequences (an ) that
converge to l according to our new version of the denition are j
M1P1 Sheet 11
x+1
2
x+5
= and lim
= 1.
x1 x + 2
x4 x + 3
3
1. Show that lim
1
= .
x1 (x 1)2
2. Prove that lim
5x 2 + 1
2x2 + 1
, g (x) =
x
x
do have a limit at x = 0.
3. Show that the functions f, g : R\cfw_0 R dened as f (x) =
do not have a limit at x =
M1P1 Sheet 10
1. If f (x) = 2x 1 for x = 3 and f (3) = 10, prove directly from the denition of a limit,
that f (x) 5 as x 3.
2. Prove, directly from the denition of a limit, that if f (x) l as x a and f (x) m
as x a, then l = m (that is, limits are unique
M1P1 Sheet 1
The aim of this preliminary sheet is to help you begin to understand the dicult denition
of an l given in the lectures. Recall that this means: given any > 0, there is a positive
integer N such that n N = |an l| < .
1. Students sometimes stat
M1P1 Sheet 2
1. For each of the following sequences (an ), guess its limit l, and prove that an l (that
is, for any given > 0, nd a positive integer N so that n N = |an l| < ).
1
(i)
n
(ii)
n+1
.
n+2
2. For each of the following sequences (an ), guess it
M1P1 Sheet 3
1. Use the theorems on sums, products and quotients of limits to nd (rigorously!) the
limits of the following sequences:
(i)
3n2 + 1
n2 + 5n
(ii)
4n3 6n + 7
2n3 + n2 + 1
(iii)
1+2+ 3+.+n
.
n2
Indicate carefully at each step which theorem you
M1P1 Sheet 4
1. Give the proof (omitted in the lectures) of the Theorem an l = |an | |l|.
[Hint: remember the general rule |a| |b| |a b|.]
2. Give the proof (omitted in the lectures) of the Theorem
an 0 if and only if |an | 0.
[Hint: write down the deniti
M1P1 Sheet 5
1. Prove that if (an ) and (bn ) are both Cauchy sequences, then so is (an + bn ).
[Hint: this is similar to the proof of the Theorem (2.4) about adding convergent sequences.]
2. Prove that if (an ) and (bn ) are both Cauchy sequences, then s
M1P1 Sheet 6
1. Prove that if |an | 1 for all n, then the sequence (an ) does not converge to zero.
[This fact was used in the lectures in the discussion of the G.P.]
2. For each of the following series
un , nd a simple formula for the sum sn of the rst
1
M1P1 Sheet 7
un :
1. Test for convergence of the following series
1
(a) un = n/2n
(c) un = (n!)1/n
(e) un = n/(n2 + 1)
1
(g) un = n
3 + 5n
(b) un = n2 /3n
(d) un = (n!)2 /(2n)!
(f) un = (n + 7)/(n3 + 1)
(h) un = 7 cos(n5/2 )/n3/2 .
2. Prove that the serie
M1P1 Sheet 8
1. Find the radius of convergence of each of the following power series:
n3 z n
(a)
1
1
z 2n+1 /(2n + 1)!
(d)
(e)
0
z 2n /(2n)!
(b)
1
(c)
1
1.2. . . . .n
zn
3.5. . . . .(2n + 1)
2. Let an be any sequence of numbers such that
1
2
radius of con
M1P1 Sheet 9
1. Give examples of power series
1
an z n with radius of convergence 1 such that
(a) the series does not converge for any z such that |z | = 1;
(b) the series converges for every z such that |z | = 1;
(c) the series converges for z = 1 and di