Assignment 4
Exercise 1 Give an example of the following, or explain why no such example can exist.
- A sequence with infinitely many terms equal to zero, but does not converge to zero.
- A sequence with infinitely many terms equal to zero, which converges to a nonzero number.
- A sequence of irrational numbers that converges to a rational number.
- A convergent sequence where every term is an integer.
- Two divergent sequences \(a_n,b_n\) where \(a_n+b_n\) is convergent.
- Two divergent sequences \(a_n,b_n\) where \(a_nb_n\) is convergent.
Solution.
This is possible: the sequence \(0,1,0,1,0,1,0,1,\ldots\) has every other term equal to zero but is divergent.
This is impossible. If \(s_n\to a\) and \(a\neq 0\), then for \(\epsilon = |a|/2\) we can find an \(N\) for which all \(n>N\) have \(|s_n-a|<\epsilon\). Unpacking this, \[a-|a|/2\leq s_n\leq a+|a|/2\] Whether \(a\) is greater than or less than zero, this interval does not contain zero. So, no terms past \(N\) can possibly be zero, and there must only be finitely many terms equal to zero (they can only be in the terms \(s_1,s_2,\ldots,s_N\))
This is possible: given any irrational \(x\) the sequence \(x/n\) converges to \(0\) by the limit theorems.
This is possible: take the constant sequence \(1,1,1,1,1,1,\ldots\)
Let \(a_n=n\) and \(b_n= -n\). Then each sequence diverges, but \(a_n+b_n= n+(-n)=0\) is the constant sequence \(0,0,0,\ldots\) which converges to \(0\).
Let \(a_n=0,1,0,1,0,1,0,\ldots\) and \(b_n=1,0,1,0,1,0,1,\ldots\). Each of these is divergent but their product \(a_nb_n\) is the zero sequence \(0,0,0,0\ldots\) so this is convergent.
Exercise 2 Prove, directly from the definition of convergence (not using limit theorems / squeeze theorem / etc), that \[\frac{2n-2}{5n+1}\to \frac{2}{5}\]
Solution. Let \(\epsilon>0\). Choose \(N=\frac{12-5\epsilon}{25\epsilon}\). Since \(\epsilon\neq0\), this fraction is defined. Now, let \(n>N\). Then we have \[\begin{align*} \frac{12-5\epsilon}{25\epsilon}& < n\\ \frac{12}{25\epsilon}-5& < n\\ \frac{12}{\epsilon}-5& < 25n\\ \frac{12}{\epsilon}& < 25n+5\\ \end{align*}\] Since \(\epsilon>0\), we can multiply both sides of this inequality by \(\epsilon\). \[\begin{align*} 12&<(25n+5)\epsilon\\ \frac{12}{25n+5}&<\epsilon \end{align*}\] Since \(n\in\NN\), we have \(\frac{12}{25n+5}>0\), so \(\left|\frac{12}{25n+5}\right|=\frac{12}{25n+5}\). By previous proof, we have \(|x|=|-x|\). Therefore, we can rewrite our inequality as \[\begin{align} \left|\frac{-12}{25n+5}\right|&<\epsilon\\ \left|\frac{10n-10-10n-2}{5(5n+1)}\right|&<\epsilon\\ \left|\frac{10n-10}{5(5n+1)}-\frac{10n+2}{5(5n+1)}\right|&<\epsilon\\ \left|\frac{2n-2}{5n+1}-\frac{2}{5}\right|&<\epsilon \end{align}\] Therefore, for any arbitrary \(\epsilon\), there exists an \(N\) such that for all \(n>N\), \(\left|\frac{2n-2}{5n+1}-\frac{2}{5}\right|<\epsilon\). Thus, by definition, \(\frac{2n-2}{5n+1}\rightarrow\frac{2}{5}\).
Exercise 3 If \(a_n\) is a convergent \(a_n\geq L\) for all \(n\), then \(\lim a_n\geq L\). Similarly prove if \(a_n\) is a convergent \(a_n\leq U\) for all \(n\), then \(\lim a_n\leq U\).
To gether these give a very useful way to bound a limit: if \(s_n\) is a convergent sequence with \(L\leq s_n\leq U\) for all \(n\), then \(L\leq \lim s_n\leq U\).
Solution. Let \(a_n\to a\) and assume for contradiction \(a<L\). Then \(L = a +\beta\) for some \(\beta>0\). Set \(\epsilon=\frac{\beta}{2}\). The interval (\(a-\epsilon,a+\epsilon\)) contains numbers strictly less then \(L\). By the definition of convergence, we know there exists an \(N\) s.t. for all \(n>N\), \(a_n\) lies in this interval. But this is a contradiction, because we know \(a_n \geq L\) for all \(n\), so we know \(a \geq L\).
A similar argument works for the other part. Let \(a_n\to b\). For the sake of contradiction, suppose \(a>U\). Then \(U = b -\beta\) for some \(\beta>0\). Then, for some \(\epsilon>0\), we can let \(\epsilon=\frac{\beta}{2}\). Thus The interval (\(a-\epsilon,a+\epsilon\)) contains numbers strictly greater then \(U\). By the definition of convergence, we know there exists an \(N\) s.t. for all \(n>N\), \(a_n\) lies in this interval. But this is a contradiction, because we know \(a_n \leq U\) for all \(n\), so we know \(a \leq U\).
Another way of expressing this idea:
Solution. First we consider the lower bound: assume for the sake of contradiction that \(a_n\to a\) but \(a<L\) even though for all \(n\) we know \(a_n\geq L\). Set \(\epsilon = \frac{L-a}{2}\): then we can find an \(N\) where for all \(n>N\) we know \(|a_n-a|<\epsilon\).
Unpacking this shows \[-\frac{L-a}{2}< a_n-a<\frac{L-a}{2}\] which implies (looking only at the upper bound, as that’s what’s relevant for us) \[a_n<a+\frac{L-a}{2}=\frac{L+a}{2}\] Thus, \(a_n\) must be less than the average of \(a\) and \(L\), but since \(a<L\) this average is less than \(L\)! This is a contradiction, since we know \(a_n\geq L\) for all \(n\).
A very similar proof works for the upper bound: if \(a_n\to a\) and \(a>U\) even though \(a_n\leq U\) for all \(n\), setting \(\epsilon = \frac{a-U}{2}\) allows us to get an \(N\) where \(|a_n-a|<\epsilon\), whose lower bound unpacks into \[a-\frac{a-U}{2}=\frac{a+U}{2}<a_n\]
Thus, \(a_n\) must be greater than the average of \(a\) and \(U\), which (since \(a\) is greater than \(U\) itself) must be greater than \(U\), contradicting the fact that \(a_n\leq U\).
Exercise 4 Let \(x_n,y_n\) be two convergent sequences with the same limit. Prove directly (ie don’t just use the squeeze theorem) that the interleaved sequence \(s_n\) defined by \[x_0,y_0,x_1,y_1,x_2,y_2,x_3,y_3,\ldots,x_n,y_m\ldots\] also converges, and has the same limit.
Solution. Call the common limit of \(x_n\) and \(y_n\) by \(L\). Choosing an \(\epsilon>0\) we know that we can find an \(N_x\) such that \(|x_n-L|<\epsilon\) for all \(n>N_x\), and an \(N_y\) such that $|y_n-L|<for all \(n>N_y\). Now let \(N=4\max\{N_x,N_y\}\) (the factor 4 is overkill and you can do better, but I’m not going to bother).
Choosing an arbitrary \(n>N\), we want to show that \(|s_n-L|<\epsilon\). Since \(s_n\) is made from a mix of terms from \(x_n\) and \(y_n\) we have two cases. First consider if \(n\) is even, so we can write \(n=2k\), and \(s_n= y_k\) as the terms are interleaved every other. Now, as \(n>N\) we see \(n>4\max{N_x,N_y}\) implies that \(k>2\max{N_x,N_y}\) and so in particular, \(k>N_y\). Thus, by our assumption on \(N_y\), we have that \(|y_k-L|<\epsilon\), so \(|s_n-L|<\epsilon\) as desired.
The reasoning is identical for \(n\) odd, where now \(n=2k+1\) and \(s_n=x_k\). Since \(n>4\max\{N_x,N_y\}\) we know \(n-1=2k>3\max\{N_x,N_y\}\), thus \(k>\max{N_x,N_y}\) so \(|x_k-L|<\epsilon\) or equivalently \(|s_n-L|<\epsilon\).
Exercise 5 Use the squeeze theorem to prove that \[\lim\left(\frac{n^3-2-\frac{1}{n^3}}{3n^3+5}\right)^{2n+7}=0\] Hint: show this sequence is positive for all large enough \(n\), and then bound it above by an easy-to-compute sequence that we can show goes to zero. Notice how much easier this is than trying to find an \(N\) for a given \(\epsilon\)!
Solution. Starting with the given fraction, two things that we can do to make a larger sequence it is to make it a smaller denominator, and a larger numerator, \[\begin{align*} \bigg( \dfrac{n^3-2-\frac{1}{n^3}}{3n^3+5}\bigg) ^{2n+7} &< \bigg( \dfrac{n^3}{3n^3}\bigg) ^{2n+7} \\ \bigg( \dfrac{n^3-2-\frac{1}{n^3}}{3n^3+5}\bigg) ^{2n+7} &< \bigg( \dfrac{1}{3} \bigg)^{2n+7} \end{align*}\] And a way to make a smaller sequence is to use the sequence \(0\). This is because, after \(n>1\) the fraction stays positive.
\[ 0 \leq \bigg( \dfrac{n^3-2-\frac{1}{n^3}}{3n^3+5}\bigg) ^{2n+7} \leq \bigg( \dfrac{1}{3} \bigg)^{2n+7} \]
Now we just need to work with the upper bound a little more: since \(2n+7>n\) we see that \(3^{2n+7}>3^n\) and
\[\left(\frac{1}{3}\right)^{2n+7}<\left(\frac{1}{3}\right)^n\]
That means we can replace our upper bound with this:
\[0\leq \bigg( \dfrac{n^3-2-\frac{1}{n^3}}{3n^3+5}\bigg) ^{2n+7} < \left(\frac{1}{3}\right)^{2n+7}<\left(\frac{1}{3}\right)^n\]
And since we have proven that \(a^n\to 0\) and if \(|a|<1\) we can apply this to \((1/3)^n\) and see it goes to zero.
\[0\leq \lim \bigg( \dfrac{n^3-2-\frac{1}{n^3}}{3n^3+5}\bigg) ^{2n+7}\leq \left(\frac{1}{3}\right)^n=0\]
So we can see that the limit must be \(0\), via the squeeze theorem.
Exercise 6 Prove that for every \(x\in\RR\) there exists a sequence \(y_n\) of irrationals with \(y_n\to x\).
Solution. Let \(x \in \mathbb{R}\) be arbitrary. We know that the sequence \(a_n = x\) has the property that \(a_n \rightarrow x\). Similarly, we know that the sequence \(b_n = \frac{1}{n}\) has the property that \(b_n \rightarrow 0\). So, by the sum law for limits, we know that lim\(\:(a_n + b_n)\) = lim\(a_n\) + lim\(b_n\) = x + 0 = x. Let’s define the sequence \(c_n\) as \(c_n = \frac{1}{n} + x\), so \(c_n \rightarrow x\).
This means that there exists a sequence \(c_n\) such that \(c_n \rightarrow x\), but note that this doesn’t require that \(c_n\) is solely irrational numbers. However, by the density of the irrationals in the real numbers (proved in an earlier homework), we know that there exists some irrational number in between any \(m, n \in \mathbb{R}\). So, lets set \(m = c_n\) and \(n = x\). According to the density of the irrationals, there must be an irrational number \(y\) such that \(x \leq y \leq c_n\) for all \(n \in \mathbb{N}\). Lets form a sequence \(y_n\), where \(y_n = y\), dictated by the inequality in the previous sentence. This sequence is made of only irrational numbers, and satisfies the equation \(a_n \leq y_n \leq c_n\), so by the squeeze theorem, \(y_n \rightarrow x\).
Exercise 7 In this problem you give an alternative proof that \(a^{1/n}\to 1\) for all \(a>0\) than the textbook’s (which used Bernoulli’s inequality). Instead, you’ll make use of the geometric sum from two assignments ago!
To recall, this stated that for any \(|r|<1\) and \(n\in\NN\), \[1+r+r^2+\cdots r^{n-1}=\frac{1-r^n}{1-r}\]
- First consider only the case that \(a>1\). Show that the geometric sum can be rewritten \(r^n-1=(r-1)(1+r+r^2+\cdots+ r^{n-1})\), and use this to prove that \(a-1 \geq n(a^{1/n}-1)\). Hint: apply it to \(r=a^{1/n}\) and do some estimating.
- Use this to show that \(0\leq a^{1/n}-1\leq \frac{a-1}{n}\) for all \(n\), and then prove (either directly, or using the squeeze theorem) that this implies \(a^{1/n}\to 1\).
- Now consider the case \(0<a<1\) and show the same. Hint: if \(0 < a < 1\) then \(1 / a >1\), and perhaps you can do a similar trick to the textbook?
Solution. We begin with algebra: clearing denominators and multiplying the geometric sum from a previous homework by \(-1\) to get in the desired form. Substituting in \(r=a^{1/n}\) to the geometric sum equality \(r^n-1=(r-1)(1+r+r^2+\cdots+ r^{n-1})\) gives \[a-1=(a^{1/n}-1)\left(1+a^{\frac{1}{n}}+a^{\frac{2}{n}}+\cdots + a^{\frac{n-1}{n}}\right)\] In the case \(a>1\), we know by previous homework that taking \(n^{th}\) roots and powers preserves inequalities, so \(a^{1/n}>1^{1/n}=1\) and so \(a^{p/n}>1^p=1\). Thus, \[1+a^{1/n}+a^{2/n}+\cdots +a^{(n-1)/n}>1+1+\cdots +1 = n\] as there are \(n\) terms in the sum. This gives the inequality we were after: \[a-1 = (a^{1/n}-1)(1++a^{1/n}+a^{2/n}+\cdots +a^{(n-1)/n}) >(a^{1/n}-1)n \]
Dividing this inequality by \(n\) gives an upper bound for \(a^{1/n}-1\): \(a^{1/n}-1< \frac{a-1}{n}\). And recalling that we already know \(a^{1/n}>1\) we see the left side of this inequality is positive so we’ve squeezed \(a^{1/n}-1\) between two known quantities: \[0< a^{1/n}-1<\frac{a-1}{n}\]
From here we prove directly that \(a^{1/n}\to 1\), though you may also choose to combine the squeeze theorem and limit laws for sums. Let \(\epsilon>0\) and choose \(N= \frac{a-1}{\epsilon}\). Then for \(n>N\) we are guaranteed \[0<a^{1/n}-1<\frac{a-1}{n}<\frac{a-1}{N}=\frac{a-1}{\frac{a-1}{\epsilon}}=\epsilon\]
and so \(|a^{1/n}-1|<\epsilon\), meaning \(\lim a^{1/n}=1\).
Next, we consider the case where \(a<1\). This means \(b=1/a\) is greater than \(1\) and so the above argument applies here: \[\lim\left(b\right)^{1/n}=1\]
Using our limit laws, we can find the limit of the reciprocal (since \(b^{1/n}\neq 0\) and the limit is nonzero)
\[\lim \frac{1}{b^{1/n}}=\frac{1}{\lim b^{1/n}}=\frac{1}{1}=1\]
But \(1/b = 1/(1/a)=a\)! Thus
\[\lim a^{1/n}=\lim \left(\frac{1}{b^{1/n}}\right)^{1/n}=\lim\frac{1}{b^{1/n}}=1\]
Exercise 8 (Limit of Products and Reciprocals) The textbook gives the scratch work of an argument for proving the following two claims:
- If \(s_n,t_n\) are convergent sequences then \(s_nt_n\) converges, and \(\lim(s_nt_n)=(\lim s_n)(\lim t_n)\)
- If \(s_n\) is a convergent sequence where \(s_n\neq 0\) and \(\lim s_n\neq 0\), then \(1/s_n\) is convergent, with \(\lim\frac{1}{s_n}=\frac{1}{\lim s_n}\).
Take the scratch work for these two theorems and rewrite as rigorous proofs (starting by choosing an \(\epsilon\), then proposing and \(N\) in terms of \(\epsilon\), and finally proving that the required inequality holds for all \(n>N\).)
Solution. This is just asking you to write rigorously (and in the right order) the sketch argument provided in the textbook, for practice.
Exercise 9 Use the limit laws to rigorously prove the following sequence converges, and to calculate its limit.
Hint: to apply the limit laws we need to work from the “inside out”: that is, before concluding the limit of a sum exists we need to first prove the limit of each of the terms does! Hint 2: think about the kind of “algebra simplifying” you do in Calc 1! Do this same algebra, to then rigorously apply limit laws together with the fact that \(\lim 1/n=0\) \[s_n=\sqrt{\frac{1}{2^n}+\sqrt{\frac{n^2-1}{3n^2-n+1}}}\]
Solution. We start with the inside most fraction \(\frac{n^2-1}{3n^2-n+1}\). Dividing the numerator and denominator both by \(n^2\) yields the equivalent fraction \[\frac{1-\frac{1}{n^2}}{3-\frac{1}{n}+\frac{1}{n^2}}\] To prove the numerator converges, we use the limit theorems for products and sums. Since \(1/n\to 0\) we know \(1/n^2 = 1/n\cdot 1/n\) converges with limit \(0\cdot 0=0\). Subtracting from the constant sequence \(1\), we see \(\lim 1-\frac{1}{n^2}=\lim 1-\lim\frac{1}{n^2}=1-0=1\). The denominator is similar: since \(1/n^2\to 0\) we see \[\lim\left(3-\frac{1}{n}+\frac{1}{n^2}\right)=\lim 3 -\lim\frac{1}{n}+\lim\frac{1}{n}\lim\frac{1}{n}=3-0+0\cdot 0 = 3\] Since additionally the denominator is never zero (as \(3n^2\geq 3n\) so \(3n^2-n\geq 2n\geq 0\) and \(3n^2-n+1\geq 1\)), the limit law for quotients applies to give \[\lim\frac{n^2-1}{3n^2-n+1}=\lim\frac{1-\frac{1}{n^2}}{3-\frac{1}{n}+\frac{1}{n^2}}=\frac{\lim\left(1-\frac{1}{n^2}\right)}{\lim\left(3-\frac{1}{n}+\frac{1}{n^2}\right)}=\frac{1}{3}\]
Now, as each term in this sequence of fractions is positive, we can apply the limit law for square roots of convergent sequences to conclude \[\lim\sqrt{\frac{n^2-1}{3n^2-n+1}}=\sqrt{\frac{1}{3}}\]
Since \(\frac{1}{2^n} = \left(\frac{1}{2}\right)^n\) we can apply our basic limit \(a^n\to 0\) for |a|<1$ to see \(\frac{1}{2^n}\to 0\), and thus by the limit law for sums
\[\lim\left\{\frac{1}{2^n}+\sqrt{\frac{n^2-1}{3n^2-n+1}}\right\}=\lim\frac{1}{2^n}+\lim\sqrt{\frac{n^2-1}{3n^2-n+1}}=0+\sqrt{\frac{1}{3}}=\sqrt{\frac{1}{3}}\]
Finally, we make one more applicaiton of the limit theorem for roots: since the limit just calculated is nonnegative, we have
\[\lim \sqrt{\frac{1}{2^n}+\sqrt{\frac{n^2-1}{3n^2-n+1}}}=\sqrt{\sqrt{\frac{1}{3}}}=\frac{1}{\sqrt[4]{3}}\]