# Sequences and series: exercise sheet 3

Exercise.

A sequence $( a n )$ is given by

$a 1 = 2$ $a n + 1 = a n + 2 a n 2$

(a) Prove that

$a n + 1 2 - 2 = 1 4 a n 2 - 2 2 a n 2$

for all $n$.

(b) Deduce that $a n > 2$ for all $n$.

(c) Using induction, show that

$a n 2 - 2 ⩽ 8 4 n$

for each $n ⩾ 1$.

(d) Using $a n 2 - 2 = ( a n - 2 ) ( a n + 2 )$ deduce that $a n → 2$ as $n → ∞$.

Exercise.

Suppose in each case that $( a n )$ is a sequence whose $n$th term is given by the expressions below. Prove that $( a n )$ converges and find the limit. If you use the continuity of any standard functions in your answer, state which function(s) you use that you know to be continuous and at what value $l ∈ ℝ$ or $( l , m ) ∈ ℝ 2$ the continuity of that(those) function(s) is required.

(a) $n 2 + 2 n + 1 2 n 2 + 3$ ; (b) $1 - 2 n - 3 n 2 2 - 3 n - 4 n 2$ ; (c) $( n + 1 ) 2 - ( n - 1 ) 2 n + 1$; (d) $3 n 2 n 2 + n + 1 + 3 n 2 + 1$.

Exercise.

Prove that the following sequences $( a n )$ do not converge, using the following method.

Suppose first that the sequence in question converges to some $l ∈ ℝ$. Then find convergent sequences $( b n )$, $( c n )$, ..., and continuous functions $f , g , …$ such that some expression involving $f , g , … , ( a n ) , ( b n ) , ( c n ) , …$ evaluates to a sequence that you in fact know to be non-convergent.

The convergence or nonconvergence of any sequences you use must be justified either by stating that they are "standard" convergent/nonconvergent sequences discussed in lectures (such as $1 n$ or $( -1 ) n$) or by giving a proof.

(a) $( -1 ) n ( 1 - 2 - n ) + 1$.

(b) $( -1 ) n ( 2 - n ) + n 2 1 + n$.

(c) $1 2 ( cos ( n ) + ( -1 ) n 1 n )$.

(d) $sin ( n ) + n cos ( n ) 1 + n$.

Exercise.

The sequence $( a n )$ is defined inductively by $a 1 = a 2 = 1$ and $a n + 2 = a n + a n + 1 + 1 4$. Prove the following by induction on $n$.

(a) $a n ⩽ 1$ for all $n$.

(b) $a n ⩾ 1 2$ for all $n$.

(c) Prove the following holds for all $n ⩾ 1$: $a n ⩾ a n + 1 ⩾ a n + 1 3$.

(d) Say whether $( a n )$ has a limit in $ℝ$ and if it does, find this limit.

Exercise.

The sequence $( c n )$ is defined by $c 1 = 1$, $c 2 = 1 2$, and $c n + 2 = 3 c n + 1 + c n 4$. Prove the following.

(a) $1 2 ⩽ c n ⩽ 1$ for all $n$.

(b) $c n + 1 - c n = ( -1 ) n 2 4 n$ for all $n$.

(c) The subsequences $( c 2 n - 1 )$ and $( c 2 n )$ are both monotonic. (Hint: use (b) above to find a formula for $c n + 2 - c n$.)

(d) Explain why these results imply that $( c 2 n - 1 )$ and $( c 2 n )$ both converge, and in fact both converge to the same limit $l ∈ ℝ$. Which theorem from the course allows you to deduce that $l$ is the limit of the original sequence $( c n )$?

(e) Find the limit $l$. Suggestion: Use (b) repeatedly to get an expression for $c n$ in terms of $n$. Sum this series then find the limit.

## Sample solutions to selected exercises

Solution to 3.2(a). Write

$n 2 + 2 n + 1 2 n 2 + 3 = 1 + 2 n + 1 n 2 2 + 3 n 2$

and observe that $1 n → 0$ and $1 n 2 → 0$ as $n → ∞$ and so by the constant sequences $1 , 2$ and the continuity of $+ , ·$, we have $1 + 2 n + 1 n 2 → 1$ as $n → ∞$. Also by the constant sequences $2 , 3$ and the continuity of $+ , ·$ again, we have $2 + 3 n 2 → 2$ as $n → ∞$. By the continuity of division at $( 1 , 2 )$ we have

$1 + 2 n + 1 n 2 2 + 3 n 2 → 1 2$

Solution to 3.3(a). Suppose $a n = ( -1 ) n ( 1 - 2 - n ) + 1$ defines a sequence converging to $l ∈ ℝ$. Consider

$( a n - 1 ) ( 1 - 2 - n )$

Now $1$ is a constant sequence, so converges to $1$. The sequence $2 - n$ converges to $0$ by a result from lectures. So by the continuity of $-$ at $( 0 , 1 )$ $1 - 2 - n → 1 - 0 = 1$ as $n → ∞$, by the continuity of $-$ at $( l , 1 )$, $( a n - 1 ) → l - 1$ as $n → ∞$ and by the continuity of division at $( l - 1 , 1 )$, the sequence above converges to $l - 1 1$. But this is impossible as

$( a n - 1 ) ( 1 - 2 - n ) = ( -1 ) n$

which is known from lectures to be nonconvergent.

Solution to 3.4. (a) Let $H ( n )$ be the statement $a n ⩽ 1$ . Then $H ( 1 )$ is true, as $a 1 = 1 ⩽ 1$. Also $H ( 2 )$ is true, as $a 2 = 1 ⩽ 1$. Now suppose $H ( 1 ) , H ( 2 ) , … H ( n ) , H ( n +1 )$ are all true, where $n ⩾ 1$ and argue as follows.

$a n + 2 = a n + a n + 1 + 1 4 ⩽ 1 + 1 + 1 4 = 3 4 ⩽ 1$

using $H ( n )$ and $H ( n +1 )$, which say that $a n ⩽ 1$ and $a n + 1 ⩽ 1$. So $H ( n +2 )$ is true. This completes the proof by induction.

(Note: If you are not happy about assuming an induction hypothesis that says that $H ( 1 ) , H ( 2 ) , … H ( n ) , H ( n +1 )$ are all true, think of this argument just given as showing that $P ( 1 )$ holds and $P ( n ) ⇒ P ( n + 1 )$ holds for $n ∈ ℕ$ where the statement $P ( n )$ is . Often induction proofs require the induction hypothesis to cover all previous cases of the induction statement.)

(b) Let $H ( n )$ be the statement $a n ⩾ 1 2$ . $H ( 1 )$ is true as $a 1 = 1 ⩾ 1 2$ and $H ( 2 )$ is true as $a 2 = 1 ⩾ 1 2$. Now suppse $H ( k )$ is true for all $1 ⩽ k ⩽ n + 1$ where $n ⩾ 1$ and argue as follows.

$a n + 2 = a n + a n + 1 + 1 4 ⩾ 1 / 2 + 1 / 2 + 1 4 = 2 4 = 1 2$

by $H ( n )$ and $H ( n +1 )$, so $H ( n +2 )$ is true. This completes the proof by induction.

(c) Let the statement $H ( n )$ be $a n ⩾ a n + 1 ⩾ a n + 1 3$, so as $a 1 = a 2 = 1$, $H ( 1 )$ says $1 ⩾ 1 ⩾ 2 3$ which is true. Now assume $H ( n )$ is true for $n ⩾ 1$ and compute:

$a n + 1 - a n + 2 = 4 a n + 1 - a n - a n + 1 - 1 4 = 3 a n + 1 - a n - 1 4 ⩾ ( a n + 1 ) - a n - 1 4 ⩾ 0$

as $a n + 1 ⩾ a n + 1 3$. Also,

$a n + 2 - a n + 1 + 1 3 = 3 ( a n + 1 + a n + 1 ) - 4 ( a n + 1 + 1 ) 12 = 3 a n - a n + 1 - 1 12 ⩾ 3 a n + 1 - a n + 1 - 1 12 ⩾ 0$

as $a n ⩾ a n + 1$ and $a n + 1 ⩾ 1 2$. This proves $H ( n +1 )$. So $H ( n )$ is true for all $n$, by induction.

(d) These results show that $( a n )$ is bounded (by (a) and (b)) and is monotonic (by (c)), and so by monotone convergence has a limit $l ∈ ℝ$. Now $( a n +1 )$ and $( a n +2 )$ are subsequences of $( a n )$ so by the theorem on subsequences converge to the same limit $l$. But $a n + 2 = a n + a n + 1 + 1 4$, so by the continuity of + and division by 4, $a n + 2 → l + l + 1 4$ as $n → ∞$. By uniqueness of limits we have $l = l + l + 1 4$, and solving this we get $l = 1 2$. Hence $l = 1 2$.
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