$$$$

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 $a_n{b}_n$ is convergent.

Solution.

• This is possible: the sequence $0,1,0,1,0,1,0,1,\dots$ has every other term equal to zero but is divergent.

• This is impossible. If $s_n\to a$ and $a\ne 0$, then for $ϵ=|a|/2$ we can find an $N$ for which all $n>N$ have $|s_n-a|<ϵ$. Unpacking this, $$a-|a|/2\le s_n\le a+|a|/2$$ 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 $s_1,s_2,\dots ,s_N$)

• 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 $1,1,1,1,1,1,\dots$

• Let $a_n=n$ and $b_n=-n$. Then each sequence diverges, but $a_n+b_n=n+(-n)=0$ is the constant sequence $0,0,0,\dots$ which converges to $0$.

• Let $a_n=0,1,0,1,0,1,0,\dots$ and $b_n=1,0,1,0,1,0,1,\dots$. Each of these is divergent but their product $a_n{b}_n$ is the zero sequence $0,0,0,0\dots$ 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 $ϵ>0$. Choose $N=\frac{12-5ϵ}{25ϵ}$. Since $ϵ\ne 0$, this fraction is defined. Now, let $n>N$. Then we have $$\begin{array}{rl}\frac{12-5ϵ}{25ϵ}& <n\\ \frac{12}{25ϵ}-5& <n\\ \frac{12}{ϵ}-5& <25n\\ \frac{12}{ϵ}& <25n+5\end{array}$$ Since $ϵ>0$, we can multiply both sides of this inequality by $ϵ$. $$\begin{array}{rl}12& <(25n+5)ϵ\\ \frac{12}{25n+5}& <ϵ\end{array}$$ Since $n\in \mathbb{N}$, 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 $$\begin{array}{rl}\left|\frac{-12}{25n+5}\right|& <ϵ\\ \left|\frac{10n-10-10n-2}{5(5n+1)}\right|& <ϵ\\ |\frac{10n-10}{5(5n+1)}-\frac{10n+2}{5(5n+1)}|& <ϵ\\ |\frac{2n-2}{5n+1}-\frac{2}{5}|& <ϵ\end{array}$$ Therefore, for any arbitrary $ϵ$, there exists an $N$ such that for all $n>N$, $|\frac{2n-2}{5n+1}-\frac{2}{5}|<ϵ$. Thus, by definition, $\frac{2n-2}{5n+1}\to \frac{2}{5}$.

Exercise 3 If $a_n$ is a convergent $a_n\ge L$ for all $n$, then $lim{a}_n\ge L$. Similarly prove if $a_n$ is a convergent $a_n\le U$ for all $n$, then $lim{a}_n\le U$.

To gether these give a very useful way to bound a limit: if $s_n$ is a convergent sequence with $L\le s_n\le U$ for all $n$, then $L\le lim{s}_n\le U$.

Solution. Let $a_n\to a$ and assume for contradiction $a<L$. Then $L=a+\beta$ for some $\beta >0$. Set $ϵ=\frac{\beta }{2}$. The interval ($a-ϵ,a+ϵ$) 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 $a_n\ge L$ for all $n$, so we know $a\ge L$.

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 $ϵ>0$, we can let $ϵ=\frac{\beta }{2}$. Thus The interval ($a-ϵ,a+ϵ$) 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 $a_n\le U$ for all $n$, so we know $a\le U$.

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 $a_n\ge L$. Set $ϵ=\frac{L-a}{2}$: then we can find an $N$ where for all $n>N$ we know $|a_n-a|<ϵ$.
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 $a_n\ge L$ for all $n$.

A very similar proof works for the upper bound: if $a_n\to a$ and $a>U$ even though $a_n\le U$ for all $n$, setting $ϵ=\frac{a-U}{2}$ allows us to get an $N$ where $|a_n-a|<ϵ$, 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 $a_n\le U$.

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 $$x_0,y_0,x_1,y_1,x_2,y_2,x_3,y_3,\dots ,x_n,y_m\dots$$ also converges, and has the same limit.

Solution. Call the common limit of $x_n$ and $y_n$ by $L$. Choosing an $ϵ>0$ we know that we can find an $N_x$ such that $|x_n-L|<ϵ$ for all $n>N_x$, and an $N_y$ such that $|y_n-L|<for all $n>N_y$. Now let $N=4max\{N_x,N_y\}$ (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 $|s_n-L|<ϵ$. 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 $n>4max{N}_x,N_y$ implies that $k>2max{N}_x,N_y$ and so in particular, $k>N_y$. Thus, by our assumption on $N_y$, we have that $|y_k-L|<ϵ$, so $|s_n-L|<ϵ$ as desired.

The reasoning is identical for $n$ odd, where now $n=2k+1$ and $s_n=x_k$. Since $n>4max\{N_x,N_y\}$ we know $n-1=2k>3max\{N_x,N_y\}$, thus $k>max{N}_x,N_y$ so $|x_k-L|<ϵ$ or equivalently $|s_n-L|<ϵ$.

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 $ϵ$!

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, $$\begin{array}{rl}({\displaystyle \frac{n^3-2-\frac{1}{n^3}}{3n^3+5}}{)}^{2n+7}& <({\displaystyle \frac{n^3}{3n^3}}{)}^{2n+7}\\ ({\displaystyle \frac{n^3-2-\frac{1}{n^3}}{3n^3+5}}{)}^{2n+7}& <({\displaystyle \frac{1}{3}}{)}^{2n+7}\end{array}$$ And a way to make a smaller sequence is to use the sequence $0$. This is because, after $n>1$ the fraction stays positive.

$$0\le ({\displaystyle \frac{n^3-2-\frac{1}{n^3}}{3n^3+5}}{)}^{2n+7}\le ({\displaystyle \frac{1}{3}}{)}^{2n+7}$$

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:

$$0\le ({\displaystyle \frac{n^3-2-\frac{1}{n^3}}{3n^3+5}}{)}^{2n+7}<{\left(\frac{1}{3}\right)}^{2n+7}<{\left(\frac{1}{3}\right)}^n$$

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.

$$0\le lim({\displaystyle \frac{n^3-2-\frac{1}{n^3}}{3n^3+5}}{)}^{2n+7}\le {\left(\frac{1}{3}\right)}^n=0$$

So we can see that the limit must be $0$, via the squeeze theorem.

Exercise 6 Prove that for every $x\in \mathbb{R}$ 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 $a_n\to x$. Similarly, we know that the sequence $b_n=\frac{1}{n}$ has the property that $b_n\to 0$. 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 $c_n\to x$.

This means that there exists a sequence $c_n$ such that $c_n\to x$, 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 $x\le y\le c_n$ 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 $a_n\le y_n\le c_n$, so by the squeeze theorem, $y_n\to x$.

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 $n\in \mathbb{N}$, $$1+r+r^2+\cdots r^{n-1}=\frac{1-r^n}{1-r}$$

• 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 $a-1\ge n(a^{1/n}-1)$. Hint: apply it to $r=a^{1/n}$ and do some estimating.
• Use this to show that $0\le a^{1/n}-1\le \frac{a-1}{n}$ 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 $$a-1=(a^{1/n}-1)(1+a^{\frac{1}{n}}+a^{\frac{2}{n}}+\cdots +a^{\frac{n-1}{n}})$$ 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 $ϵ>0$ and choose $N=\frac{a-1}{ϵ}$. Then for $n>N$ we are guaranteed $$0<a^{1/n}-1<\frac{a-1}{n}<\frac{a-1}{N}=\frac{a-1}{\frac{a-1}{ϵ}}=ϵ$$

and so $|a^{1/n}-1|<ϵ$, 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 $b^{1/n}\ne 0$ 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 $s_n{t}_n$ converges, and $lim(s_n{t}_n)=(lim{s}_n)(lim{t}_n)$
• If $s_n$ is a convergent sequence where $s_n\ne 0$ and $lim{s}_n\ne 0$, 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 $ϵ$, then proposing and $N$ in terms of $ϵ$, 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 $lim1/n=0$ $$s_n=\sqrt{\frac{1}{2^n}+\sqrt{\frac{n^2-1}{3n^2-n+1}}}$$

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 $lim1-\frac{1}{n^2}=lim1-lim\frac{1}{n^2}=1-0=1$. The denominator is similar: since $1/n^2\to 0$ we see $$lim(3-\frac{1}{n}+\frac{1}{n^2})=lim3-lim\frac{1}{n}+lim\frac{1}{n}lim\frac{1}{n}=3-0+0\cdot 0=3$$ Since additionally the denominator is never zero (as $3n^2\ge 3n$ so $3n^2-n\ge 2n\ge 0$ and $3n^2-n+1\ge 1$), the limit law for quotients applies to give $$lim\frac{n^2-1}{3n^2-n+1}=lim\frac{1-\frac{1}{n^2}}{3-\frac{1}{n}+\frac{1}{n^2}}=\frac{lim(1-\frac{1}{n^2})}{lim(3-\frac{1}{n}+\frac{1}{n^2})}=\frac{1}{3}$$

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

$$lim\{\frac{1}{2^n}+\sqrt{\frac{n^2-1}{3n^2-n+1}}\}=lim\frac{1}{2^n}+lim\sqrt{\frac{n^2-1}{3n^2-n+1}}=0+\sqrt{\frac{1}{3}}=\sqrt{\frac{1}{3}}$$

Finally, we make one more applicaiton of the limit theorem for roots: since the limit just calculated is nonnegative, we have

$$lim\sqrt{\frac{1}{2^n}+\sqrt{\frac{n^2-1}{3n^2-n+1}}}=\sqrt{\sqrt{\frac{1}{3}}}=\frac{1}{\sqrt[4]{3}}$$