# Limits of series proofs help needed

Maybe because of the reason that I am not confident with series, I could not explicitly prove the following theorems. Could you please give me a hand?

1) Let $x_n$ ($n$ is from $1$ to $\infty$) and $y_n$ ($n$ is from $1$ to $\infty$) be sequences of real numbers such that $x_n \to a$, $y_n \to b \neq 0$, $y_n \neq 0$ for all $n \in \mathbb{N}$. Prove that $x_n/y_n \to a/b$.

For this question, I know that if $x_n$ and $y_n$ were functions, then I could start by showing $\lim y(x) \to 1/b$ and then carry on by the using $\lim (f(x) \cdot f(y)) = \lim f(x) \cdot \lim f(y)$. But now I do not have an idea because $x_n$ and $y_n$ are series.

2) Let $(a_n)$ and $(b_n)$ be sequences of real numbers such that $a_n \leq b_n$ for all $n \in \mathbb{N}$. Prove that if $a_n \to a$ and $b_n \to b$, then $a \leq b$.

This is so obvious but I do not have an idea how to prove it.

3) We say that a sequence of real numbers $(a_n)$ goes to $−\infty$ and write $a_n \to -\infty$ if for every $C < 0$ there is an $N$ such that $a_n < C$ for $n > N$.

a) Prove that if $a_n \to -\infty$, $a_n \neq 0$, then $1/a_n \to 0$.

b) Prove that if $b_n \to 0$, $b_n < 0$, then $1/b_n \to -\infty$.

I do not have an idea where to start. Your help will be a life saver.

Regards

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I’ll do a very careful job of proving the first one; see if that’s enough for you to go on and do the others on your own.

To show that $\dfrac{x_n}{y_n}\to\dfrac{a}b$, we must show that for every $\epsilon>0$ there is an $N_\epsilon\in\Bbb N$ such that $$\left|\frac{x_n}{y_n}-\frac{a}b\right|<\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;,$$ so start by fixing $\epsilon>0$.

We know that by taking $n$ large enough we can get $x_n$ very close to $a$ and $y_n$ very close to $b$; how close do we need them to get in order to ensure that $$\left|\frac{x_n}{y_n}-\frac{a}b\right|<\epsilon\;?$$

A little algebra is in order: $$\left|\frac{x_n}{y_n}-\frac{a}b\right|=\left|\frac{bx_n-ay_n}{by_n}\right|=\frac{|bx_n-ay_n|}{|by_n|}\;.$$ Now $b\ne 0$, so there is an $M\in\Bbb N$ such that $|y_n|>\frac12|b|>0$ for all $n\ge M$, and for all $n\ge M$ we then have $$\frac{|bx_n-ay_n|}{|by_n|}<\frac{|bx_n-ay_n|}{\frac12b^2}=\frac2{b^2}|bx_n-ay_n|\;.$$ If we can find an $M_\epsilon\in\Bbb N$ such that $$|bx_n-ay_n|<\frac{b^2}2\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;,$$ we’re done: we let $N_\epsilon=\max\{M,M_\epsilon\}$, and for each $n\ge N_\epsilon$ we have

$$\left|\frac{x_n}{y_n}-\frac{a}b\right|=\frac{|bx_n-ay_n|}{|by_n|}<\frac2{b^2}|bx_n-ay_n|<\frac2{b^2}\cdot\frac{b^2}2\epsilon=\epsilon\;,$$ as desired.

Now we use the trick of subtracting and adding the same thing and apply the triangle inequality:

\begin{align*} |bx_n-ay_n|&=|bx_n-ba+ab-ay_n|\\ &\le|bx_n-ba|+|ab-ay_n|\\ &=|b||x_n-a|+|a||b-y_n| \end{align*}

Since $\langle x_n:n\in\Bbb N\rangle$ converges to $a$, we know that there is a $K_1\in\Bbb N$ such that $|x_n-a|<\dfrac{|b|\epsilon}4$ whenever $n\ge K_1$, and this ensures that $|b||x_n-a|<\dfrac{b^2\epsilon}4$ whenever $n\ge K_1$. If $a=0$, that already ensures that $$|bx_n-ay_n|\le|b||x_n-a|+|a||b-y_n|=|b||x_n|<\frac{b^2\epsilon}4<\frac{b^2\epsilon}2\;,$$ and we don’t have to worry about the $|a||b-y_n|$ term. If $a\ne 0$, the fact that $\langle y_n:n\in\Bbb N\rangle$ converges to $b$ ensures that there is a $K_2\in\Bbb N$ such that $|y_n-b|<\dfrac{b^2\epsilon}{4|a|}$ for all $n\ge K_2$. Thus, $$|bx_n-ay_n|<|b||x_n-a|+|a||b-y_n|<\frac{b^2\epsilon}4+|a|\frac{b^2\epsilon}{4|a|}=\frac{b^2\epsilon}2$$ whenever $n\ge\max\{K_1,K_2\}$.

If $a=0$ let $M_\epsilon=K_1$, and if $a\ne 0$ let $M_\epsilon=\max\{K_1,K_2\}$; then for all $n\ge K$ we have $$|bx_n-ay_n|<\frac{b^2\epsilon}2\;,$$ which is exactly what we needed: with $N_\epsilon=\max\{M,M_\epsilon\}$ we now know that $$\left|\frac{x_n}{y_n}-\frac{a}b\right|<\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;,$$ which is exactly what it means to say that $$\left\langle\frac{x_n}{y_n}:n\in\Bbb N\right\rangle$$ converges to $\dfrac{a}b$.

Added: I’ll do part (b) of (3). The hypotheses are that $b_n<0$ for each $n\in\Bbb N$ and that $\lim\limits_{n\to\infty}b_n=0$, and we want to show that $\lim\limits_{n\to\infty}\frac1{b_n}=-\infty$. The first step is to write down exactly what this means: we’re trying to show that for each $C<0$ there is an $N_C$ such that $\frac1{b_n}<C$ whenever $n\ge N_C$, so start by fixing some $C<0$.

Now look at what you have to work with: $\lim\limits_{n\to\infty}b_n=0$, so for each $\epsilon>0$ there is an $N_\epsilon\in\Bbb N$ such that $|b_n-0|<\epsilon$ whenever $n\ge N_\epsilon$. You also know that each $b_n$ is negative, so you can rewrite this as follows:

for each $\epsilon>0$ there is an $N_\epsilon\in\Bbb N$ such that $-\epsilon<b_n<0$ whenever $n\ge N_\epsilon$.

If $-\epsilon<b_n<0$, then $0<-b_n<\epsilon$, so $\dfrac1{-b_n}>\dfrac1\epsilon$, and $\dfrac1{b_n}<-\dfrac1\epsilon$. Thus, for any $\epsilon>0$ we know that $$\frac1{b_n}<-\frac1\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;.\tag{1}$$ We’re trying to fine an $N_C$ such that $$\frac1{b_n}<C\quad\text{whenever}\quad n\ge N_C\;.\tag{2}$$ $(1)$ and $(2)$ look very similar. In fact, if we let $\epsilon=-\frac1C$ and then set $N_C=N_\epsilon$, they become identical.

So that’s what we do: given $C<0$, let $\epsilon=-\frac1C$. By hypothesis there is an $N_\epsilon\in\Bbb N$ such that $(1)$ is true. Now let $N_C=N_\epsilon$, and we see that $(2)$ is true: we’ve found the needed $N_C$. This is exactly what’s required in order to conclude that $\lim\limits_{n\to\infty}\frac1{b_n}=-\infty$.

Added2: For part (a) of (3) we’re assuming that $\lim\limits_{n\to\infty}a_n=-\infty$, where $a_n\ne 0$ for $n\in\Bbb N$, and we want to prove that $\lim\limits_{n\to\infty}\frac1{a_n}=0$. In order to do this, we must show that for each $\epsilon>0$ there is an $N_\epsilon\in\Bbb N$ such that $$\left|\frac1{a_n}-0\right|<\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;.\tag{3}$$ We can simplify our goal a little by rewriting $(3)$: we want to show that for each $\epsilon>0$ there is an $N_\epsilon\in\Bbb N$ such that $$\frac1{|a_n|}<\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;.\tag{4}$$

What do we have to work with? We know that $\lim\limits_{n\to\infty}a_n=-\infty$, which by definition means that for each $C<0$ there is an $N_C\in\Bbb N$ such that

$$a_n<C\quad\text{whenever}\quad n\ge N_C\tag{5}\;.$$ Now stop and think about what $(4)$ and $(5)$ say. $(5)$ says that we can make $a_n$ a negative number of large absolute value by taking $n$ large enough; that’s what we know that we can do. $(4)$, says that we can make $\frac1{|a_n|}$ very small by taking $n$ large enough; that’s what we want to be able to do. But making $a_n$ a negative number of large magnitude automatically makes $\frac1{|a_n|}$ a very small number, so if we can do the one, we can do the other. All we have to do is write down the details.

Let $\epsilon>0$ be any positive real number, and let $C=-\frac1\epsilon$. Then $C<0$, so by hypothesis there is an $N_C\in\Bbb N$ such that $(5)$ holds. Let $N_\epsilon=N_C$. Then for each $n\ge N_\epsilon$ we know that$n\ge N_C$, so $$a_n<C=-\frac1\epsilon\;.\tag{6}$$ Multiply $(6)$ by $-1$: $$-a_n>-C=\frac1\epsilon\;.$$ These are positive numbers, so taking reciprocals yields $-\dfrac1{a_n}<\epsilon$ for each $n\ge N_\epsilon$. Finally, $a_n$ is negative, so $-a_n=|a_n|$, and we’ve just shown $(4)$: that $\dfrac1{|a_n|}<\epsilon$ whenever $n\ge N_\epsilon$. This is precisely what is meant by the statement that $\lim\limits_{n\to\infty}\frac1{a_n}=-\infty$, so we’re done.

All of these arguments have worked the same way. I’ve used the definition of limit to write down exactly what I need to show and exactly what information I’m given. I’ve then tried to see how the two are related. In (3) it’s basically just been a matter of recognizing that when $|x|$ gets large, $\frac1x$ gets close to $0$, and vice versa. In (1) I actually had to work quite a bit harder, because the connection between what I was given and what I wanted was much less obvious. That’s where a certain amount of experience and practice are helpful: that trick of subtracting and adding the same intermediate quantity and then using the triangle inequality is very common, but it takes some ingenuity to discover it if you’ve never seen it before.

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@BrianMScott thank you so much for your help. Here is the situation: I think I could successfully prove the problem number 2 by using properties: if an>0 then lim(an)>0 and lim(an±bn)=lim(an)±lim(bn)=a±b. However still stuck at problem number 3. I think I have problems with using the limit's definition. Could you please help me for the 3rd question also? I really appreciate your help. –  Amadeus Bachmann Oct 14 '12 at 18:02
@Zxy: I’ve expanded my answer to address (3). –  Brian M. Scott Oct 14 '12 at 22:20
@BrianMScott I am so much thankful for your help. However I could not prove part (a) of 3 too. :/ We know that an<C<0 (1/an>1/C) whenever n≥Nc and want to fix ϵ>0 to show that ∣(1/an)-L∣<ϵ whenever n≥Nϵ and find L=0. So, -ϵ<(1/an)-L<ϵ and -1/ϵ>(1-anL)/an>1/ϵ however I could not figure out how to continue from here. Maybe I am doing wrong to try to find that L=0 with algebraic operations, maybe it is something I should decide after I find some relations. I am sorry for the inconvenience. :/ –  Amadeus Bachmann Oct 15 '12 at 7:18
@BrianMScott this looks better: -ϵ+L < 1/an < ϵ+L , 1/(-ϵ+L) > an > 1/ϵ+L whenever n≥Nϵ and an<C<0 whenever n≥Nc. But I could not figure out what to do next. –  Amadeus Bachmann Oct 15 '12 at 7:23

Remember the definition of the limit: $b$ is the limit of $(b_n)$ if for any $\delta>0$ the re is an $n_\delta$ such that for all $n\geq n_\delta:\,b_n\in(b-\delta,b+\delta)$. On 2.) Let $a_n\leq b_n$, that is $a_n+\epsilon <b_n$ for some $\epsilon>0$. However, since $a$ is the limit of $(a_n)$, for any $\delta>0$ and hence also for our $\epsilon$ there is an $n_\epsilon$ large enough such that for all $n\geq n_\epsilon:\, a_n\in(a-\epsilon,a+\epsilon)$. Thus $b_n$ is not in this interval for $n\geq \epsilon'$ and hence $b$ is neither.

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Sorry, I meant $b_n$ is not in this interval for $n\geq n_\epsilon$, of course. –  kelu Oct 14 '12 at 14:12
do you have an idea for the 3rd one? –  Amadeus Bachmann Oct 14 '12 at 21:14