# Sequences and series: revision exercise sheet

This exercise sheet contains some additional questions that may be useful for revision or review of the material in these web pages.

This is work in progress. Please return here later.

Exercise.

Using the theorem on subsequences, and/or the theorem on boundedness of convergent sequences, and/or the theorem bounding the limit of a bounded sequence, prove the following sequences $( a n )$ with $n$th term as given do not converge.

(a) $a n = 0$ if $n$ is prime, $a n = 1$ otherwise.

(b) $a n = 3 n + ( -1 ) n$.

(c) $a n = sin ( n π + π 4 )$.

(d) $a n = sin n$. (Hint: consider the subsequences $a f ( n )$ and $a g ( n )$ where $f ( n ) = [ 2 n π + π 2 ]$ and $g ( n ) = [ 2 n π + 3 π 2 ]$.)

Exercise.

Prove that $log ( n ) log ( n ) > n 2$ for sufficiently large $n$. (Take logs of both sides. What properties of the $log$ function do you require?)

Hence show that the series $∑ n = 2 ∞ 1 log ( n ) log ( n )$ converges.

Exercise.

A function $F ( x )$ is defined by setting $F ( x ) = lim n → ∞ x x n - n x n + n$.

Prove that $F ( x )$ is defined for all $x ∈ ℝ$ (i.e., the limit above exists for all $x ∈ ℝ$) and takes the following values: $F ( x ) = x$ if $| x | > 1$; and $F ( x ) = - x$ if $| x | ⩽ 1$.

(Hint: Split into cases and divide top and bottom of the fraction by appropriate numbers. You may find it helpful to use that fact that $n y n → 0$ if $| y | < 1$.)

Solution to Exercise 2, in the special case when $0 < x < 1$: Note that for these values of $x$ we have $x x n - n x n + n = x 1 n x n - 1 1 n x n + 1 → x · 0 - 1 0 + 1 = - x$ as $n → ∞$ by standard null sequences $x n$, $1 n$ and continuity of the arithmetic operations. Thus $F ( x ) = - x$ for $0 < x < 1$.

The next two exercises concern some common misconceptions and errors in beginning analysis. Many of these examples are worth learning. Some are easy but some are rather subtle and tricky. (One or two are dreadful howlers that I hope no-one reading this would ever consider writing.) If you have a good feel for things like this you are well on the way to being rather good at analysis.

Exercise.

The following statements are all false. In each case, find a counterexample.

(a) If the sequence $( a n )$ diverges and the sequence $( b n )$ diverges then their sum $( a n + b n )$ diverges.

(b) If the sequence $( a n )$ diverges and the sequence $( b n )$ diverges then their product $( a n · b n )$ diverges.

(c) If the sequence $( a n )$ is a null sequence then $∑ n = 1 ∞ a n$ converges.

(d) If $a n > B$ for all $n ∈ ℕ$ and $a n → l$ as $n → ∞$ then $l > B$.

(e) If $( a n )$ is a bounded sequence and has a convergent subsequence $a f ( n ) → l$ as $n → ∞$ then $a n → l$.

(f) If $∑ n = 1 ∞ a n = l$ and $∑ n = 1 ∞ b n = m$ both converge, then $∑ n = 1 ∞ ( a n b n )$ converges and equals $l m$.

(g) If $( a n )$ is a null sequence then $n · a n → 0$ as $n → ∞$.

(h) If the sequence $( a n )$ does not converge then $1 a n → 0$ as $n → ∞$.

(i) If $( a n )$ and $( b n )$ are both sequences and $a n b n → 0$ as $n → ∞$ then either $a n → 0$ or $b n → 0$ as $n → ∞$.

(j) If $a n b n → 0$ as $n → ∞$ and $∑ n = 1 ∞ b n$ converges, then $∑ n = 1 ∞ a n$ converges. (Hint: consider an appropriate conditionally convergent series.)

(k) If $a n b n → l ≠ 0$ as $n → ∞$ and $∑ n = 1 ∞ b n$ converges, then $∑ n = 1 ∞ a n$ converges. (Hint: let $a 2 n = b 2 n = - b 2 n + 1 = 1 log n$ and $a 2 n + 1 = 1 n - 1 log n$.)

Solution to Exercise 3(a): Let the sequence $( a n )$ be defined by $a n = ( -1 ) n$ and $( b n )$ be defined by $b n = ( -1 ) n + 1$. Then both these sequences diverge, and yet their sum is $a n + b n = -1 + 1$ or $1 + -1$ which defines the constant sequence with value 0, which converges.

## Sample solutions to selected parts

Solution to 1(a): Let $( b n )$ be the subsequence of $( a n )$ consisting of all those $a n$ with $n$ prime, and let $( c n )$ be the subsequence of $( a n )$ consisting of all those $a n$ with $n$ non-prime. Both $( b n )$ and $( c n )$ are valid subsequences as there are both infinitely many primes in $ℕ$ and infinitely many non-primes in $ℕ$. But $( b n )$ is the constant sequence with value $0$ and $( c n )$ is the constant sequence with value $1$.

Therefore $( a n )$ does not converge, for if $a n → l$ as $n → ∞$ then $b n → l$ and $c n → l$ by the theorem on subsequences, so $l = 0$ by the theorem on uniqueness of limits as $b n → 0$ since constant sequences converge to their value, and $l = 1$ by the theorem on uniqueness of limits as $c n → 1$ as this is a constant sequence also. But this means $0 = 1$, which is false.