(8.2 b,e) For each of the following sequences, determine its limit (usual Calculus reasoning will be helpful here), then prove that the sequence does indeed converge to this limit.
Let \(\epsilon > 0\). Consider \(N = \frac 16 \left( \frac {106}{3\epsilon} - 7 \right)\). Then for \(n > N\),
\[\left\vert \frac {7n-19}{3n+7} - \frac 73 \right\vert = \frac {106}{3(3n + 7)} < \frac {106}{3(3N + 7)} = \epsilon.\]So \(\lim a_n = 7/3\).
Let \(\epsilon > 0\). Consider \(N = 1/\epsilon\) and \(n > N\). Then
\[\left\vert \frac {\sin n}n \right\vert = \frac{\vert \sin n \vert}{\vert n \vert} \le \frac 1{\vert n \vert} = \frac 1n < \frac 1N = \epsilon.\]So \(\lim (\sin n)/n = 0\).
(8.6) Let \((s_n)\) be a sequence in \(\mathbb R\).
Prove that \(\lim{s_n} = 0\) if and only if \(\lim{ | s_n | } = 0\). (Hint: This is an “if and only if” statement, so you will need to prove both directions of the statement.)
(\(\Rightarrow\)) Suppose \(\lim s_n = 0\). Consider \(\epsilon > 0\). Then there exists \(N\) so that \(n > N\) implies that \(\vert s_n \vert < \epsilon\).
Observe that \(\vert \vert s_n \vert - 0 \vert = \vert s_n \vert\) which is less than \(\epsilon\) by above. So \(\lim \vert s_n \vert = 0\).
(\(\Leftarrow\)) The reverse direction is very similar to the forward direction.
Prove that if \(s_n = (-1)^n\) that \(\lim{|s_n|}\) exists but \(s_n\) diverges.
We have that \(s_n\) diverges by Example 4 in Section 8 (pp41-42). Also, \(\vert s_n \vert = 1\) which converges to \(1\) as it is constant.
(7.4) Give examples of
A sequence of irrational numbers converging to a rational number
For example, let \(s_n = \pi/2^n\).
A sequence of rational numbers converging to an irrational number
For example, let \(s_n = \sum_{k=0}^n{1/(k!)}\) which converges to \(e\).
For each of the following false statements, give a counterexample.
If \(s_n\) and \(t_n\) are two divergent sequences, their sum \(s_n + t_n\) also diverges.
For example, let \(s_n\) diverge and let \(t_n = -s_n\).
If a sequence \(s_n\) converges, so too does the sequence \(t_n = \sum_{i=1}^n{s_i}\).
For example, let \(s_n = 1/n\).
If the sequence \(s_n\) converges and the sequence \(t_n\) diverges, the product \(s_nt_n\) also diverges.
For example, let \(t_n\) be any divergent sequence and let \(s_n = 0\).
Another example, let \(t_n\) be any divergent, nonzero sequence and let \(s_n = 1/t_n\).
(9.12) Suppose \(s_n \ne 0\) for all \(n \in \mathbb N\) and that the limit \(\lim {\left|\frac{s_{n+1}}{s_n}\right|} = L\) exists.
Show that if \(L < 1\) then \(\lim s_n = 0\).
Proof. As \(L < 1\) there exists a number \(L < a < 1\) by the Completeness Axiom. Consider \(\epsilon = a - L\) which is positive as \(L < a\).
Claim: There exists \(N\) such that for all \(n \ge N\), \(\vert s_{n+1}\vert < a \vert s_n\vert\). Let \(N \in \mathbb N\) be that for which
\[\left\vert \left\vert \frac {s_{n+1}}{s_n} \right\vert - L \right\vert < \epsilon\]for all \(n \ge N\), which exists as \(\lim {\left\vert \frac{s_{n+1}}{s_n}\right\vert} = L\). (Notice: For simplicity’s sake, we’re taking \(N\) here so that even \(n = N\) satisfies the convergence inequality. We may do this, as once we know \(N\) with the strict inequality exists, we can just choose the next natural number larger by the Archimedean Property.) So, for such \(n\) we have that
\[L - \epsilon < \left\vert \frac {s_{n+1}}{s_n} \right\vert < L + \epsilon = a.\]Notice that the right part of this inequality implies that \(\vert s_{n+1} \vert < a \vert s_n \vert\), as claimed.
Claim: For all natural numbers \(k\), \(\vert s_{N+k}\vert < a^k \vert s_N\vert\). We prove this using induction on \(k\). The base case of \(k=1\) we have already shown, by taking \(n = N\) in the inequality \(\vert s_{n+1} \vert < a \vert s_n \vert\) which is true by choice of \(N\).
Suppose now that \(\vert s_{N+k} \vert < a^k \vert s_N \vert\). We have that \(N+k \ge N\), so we know that \(\vert s_{(N+k)+1} \vert < a \vert s_{N+k} \vert\). By our inductive hypothesis, we have that
\[\vert s_{(N+k)+1} \vert < a \vert s_{N+k} \vert < a (a^k \vert s_N \vert) = a^{k+1}\vert s_N \vert,\]so by induction the claim is true.
Claim: \(\lim s_n = 0\). Observe that we have now that \(0 < \vert s_{N+k} \vert < a^k \vert s_N \vert\). Consider \(n > N\); then consider \(k = n-N\), which is a natural number as \(N \in \mathbb N\) and \(n > N\). So
\[0 < \vert s_{n} \vert < a^n \left\vert \frac {s_N}{a^N} \right\vert\]Notice we’re actually considering the sequence \((\vert s_n \vert)_{n=N}^{\infty}\) here. But think about why if we can prove the limit for this sequence, we’ve also proved the limit for the full sequence as well.
The LHS has limit \(0\), and the RHS has limit \(0\) (why?). So by the squeeze theorem, the sequence \((\vert s_n \vert)_{n=N}^{\infty}\) has limit 0 and so \(\lim \vert s_n \vert = 0\) as well. By question (2a), this means \(\lim s_n = 0\). \(\Box\)
Show that if \(L > 1\) then \(\lim | s_n | = +\infty\).
Consider the sequence \(t_n = 1/s_n\). So
\[\lim \left\vert \frac{t_{n+1}}{t_n} \right\vert = \lim \left\vert \frac{s_{n}}{s_{n+1}} \right\vert = 1/L < 1.\]So \(\lim t_n = 0\) and by question 2a, \(\lim \vert t_n \vert = 0\). As \(\vert s_n \vert\) is a positive sequence and \(\lim 1/\vert s_n \vert = \lim \vert t_n \vert = 0\) we have that \(\lim \vert s_n \vert = +\infty\).