Sequences and series: exercise sheet 4

Exercise.

Consider the sequence defined by a 1 = 1 and a n + 1 = 1 + a n .

(a) Assuming for the moment that ( a n ) is convergent, by solving a quadratic equation or otherwise make a reasonable conjecture for a value l that might be the limit of the sequence.

(b) Prove that 1 a n < l holds for all n , by induction. Your proof must not use the conjecture from (a) that a n l , only the value of l itself.

(c) Using part (b), or by induction, prove that the sequence is monotonic nondecreasing.

(d) Which theorem from lectures allows you to deduce from (b) and (c) that a n has a limit?

(e) Apply the continuity of the function f ( x ) = 1 + x (considered as a function of positive x into the set of positive reals) to prove that the limit of the sequence ( a n ) is l , as conjectured in part (a).

Exercise.

Find supA and infA where A = { 1 n - 1 k : n , k , 1 < n < k } .

Exercise.

For the following series series n = 1 a n , determine if the series converges and if so find the limit. (Hint: use partial fractions to express a n .)

(a) a n = 1 n ( n + 1 )

(b) a n = 1 n 2 + 2 n

(c) a n = 1 n ( n 2 - 1 ) (As a 1 is not defined consider n = 2 a n .)

Exercise.

(a) Let p > 0 in and n . Prove that ( 1 + p ) n 1 + n p + n ( n - 1 ) 2 p 2 . (Use induction on n .)

(b) If r with 0 < r < 1 , show that n r n 0 as n . (Hint: write r = 1 1 + p and use (a).)

(c) Let a n = n r n where 0 < r < 1 . Show that k = 1 n a k = r ( 1 - r ) 2 ( 1 - ( n + 1 ) r n + n r n + 1 ) (Use induction on n .)

(d) Hence show that the series n = 1 n r n converges if 0 < r < 1 , and find its limit.

(e) Show that n = 1 n r n diverges if r 1 .

Standard results and theorems, including the squeeze rule, may be used if quoted correctly.

Sample solutions to selected exercises

Exercise 4.3(a)

Let s n = k = 1 n a n .

Note that a n = 1 n ( n + 1 ) = 1 n - 1 n + 1 so s n = 1 1 - 1 2 + 1 2 - 1 3 + 1 3 - 1 4 + + 1 n - 1 n + 1 hence s n = 1 1 - 1 n + 1 for all n . But 1 n + 1 0 as n (being a subsequence of a standard null sequence 1 n ) and so s n 1 - 0 = 1 by continuity of - and therefore the series converges with n = 1 a n = 1 .

Exercise 4.4(b)

Since 0 < r < 1 there is a real number p such that r = 1 1 + p and p > 0 . Now note that 0 n r n = n ( 1 + p ) n n 1 + n p + n ( n - 1 ) 2 p 2 by part (a). Simplifying this we have 0 n r n 2 n 2 n 2 + 2 p n + p 2 - p 2 n . The sequence ( b n ) on the right hand side of this inequality converges to 0 by the standard null sequence 1 n and continuity of addition, subtraction, multiplication, and of division at ( 0 , p 2 ) . (Note that p 2 0 .) Thus by the squeeze rule applied to 0 n r n b n 0 we have n r n 0 as n as required.