# 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 $A = { 1 n - 1 k : n , k ∈ ℕ , 1 < n < k }$.

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 $ℝ$ and $n ∈ ℕ$. Prove that $( 1 + p ) n ⩾ 1 + n p + n ( n - 1 ) 2 p 2$. (Use induction on $n$.)

(b) If $r ∈ ℝ$ 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.