Sequences and series: exercise sheet 2

Exercise.

The sequence ( a n ) is defined by a 0 = 2 and a n + 1 = 1 + a n 2 . Prove by induction on n that a n = 1 + 1 2 n for all n and hence show that a n 1 as n .

Exercise.

Let a 0 = 1 , a 1 = 1 , and for n = { 0 , 1 , 2 , 3 , } , let a n + 2 = 1 + a n + a n +1 7 .

(a) Show that

a n + 2 - 1 5 = 1 7 a n - 1 5 + a n + 1 - 1 5

Hence, using the induction hypothesis, H ( n ) :

k   k n a k > 1 5

show that a n > 1 5 for all n .

(b) Use (a) above and an additional induction on n to show that

0 < a n - 1 5 < 2 1 2 n

for all n .

(c) Since 1 2 n 0 as n (from elsewhere in the webpages) we have

ε > 0   N   n N   1 2 n < ε

Use this together with (b) to prove directly from the definition of convergence for ( a n ) that a n 1 5 as n .

Exercise.

Either by quoting theorems on boundedness, subsequences, uniqueness of limits, or anything else from the course so far, or by arguing directly from the definition, prove that the following sequences do not converge to any limit.

a n = 2 n ; b n = ( -1 ) n 10000000 + 1 n ; c n = sin 3 n π 4 ; d n = sin 2 n .