Real number sequences and limits.

Question

Could anyone help me? I have to solve this problem, justifying it:

$\{a_n\}$ and $\{b_n\}$ are sequences of real numbers.

Given that: $\displaystyle\lim_{n\rightarrow\infty} a_n$ = a > 0 and $\displaystyle\lim_{n\rightarrow\infty} b_n$ = $\infty$,

Show that $\displaystyle\lim_{n\rightarrow\infty} a_nb_n = \infty$.

Answer 1

First, write out the $\epsilon$-$\delta$ meanings of each limit.

$\lim_{n\to \infty}a_n=a$ means that for all $\epsilon > 0$, there exists a $N \in \Bbb{N}$ such that for all $n > N$, $|a_n-a| < \epsilon$. ANother way of writing this is $a-\epsilon < a_n < a+\epsilon$.

$\lim_{n\to \infty}b_n=+\infty$ means that for all $O \in \Bbb{R}$, there exists a $N \in \Bbb{N}$ such that for all $n > N$, $b_n > O$.

Now, we want to prove $\lim_{n\to \infty}a_nb_n=+\infty$. This means we need to show that for all $M\in \Bbb{R}$, there exists a $N \in \Bbb{N}$ such that for all $n > N$, $a_nb_n > M$.

(Note that I use $O$ in the second limit and $M$ in the third limit just so they have different variable names. This will make things less confusing later on.)

Now, choose some arbitrary $M \in \Bbb{R}$. For $a_nb_n > M$ to be true, we need $b_n > \frac{M}{a_n}$.

At this point, I have given you the basic setup of the problem, so take a moment to pause and see if you can figure the rest of it out or at least get a sense for what all of the limit statements above really mean and how the limit statement that we are trying to prove is related to the first two given limit statements. This will help you better understand the rest of the solution.

---

Now, the crucial thing to realize here is that even though we don't know $\frac{M}{a_n}$ exactly, we can figure out an upper bound for it using the first limit definition by choosing some arbitrary $\epsilon > 0$. In this case, to be safe, we will choose $\epsilon=\frac{a}{2}$. That way, we know for some $N_\epsilon$, for any $n > N_\epsilon$:

$$a-\epsilon < a_n < a+\epsilon \rightarrow \frac{a}{2} < a_n < \frac{3a}{2}$$

Now, since $a > 0$, we have $a_n$ within some safe positive range. This way, we know $a_n$ is not arbitrarily close to $0$, which means $\frac{M}{a_n}$ is not going to be an arbitrary big number, which is what we want in order to figure out an upper bound for $\frac{M}{a_n}$. At this point, we can divide $M$ by all sides of the inequality and flip the inequality signs to figure out an upper bound:

$$\frac{M}{\frac a 2} > \frac{M}{a_n} > \frac{M}{\frac{3a}{2}} \rightarrow \frac{2M}{a} > \frac{M}{a_n} > \frac{2M}{3a}$$

Now, we know that the upper bound for $\frac{M}{a_n}$ is $\frac{2M}{a}$, so if we know that $b_n > \frac{2M}{a}$, then we know that $b_n > \frac{M}{a_n}$ by the transitive property, so we will focus on showing $b_n > \frac{2M}{a}$.

Clearly, $\frac{2M}{a}$ is a constant real number (i.e. it does not depend on the value of $n$), so we can now use the second limit to help show this inequality. From the second limit, we know that if we choose $O=\frac{2M}{a}$, then there exists an $N_O \in \Bbb{N}$ such that for all $n > N_O$, $b_n > \frac{2M}{a}$. This is what we wanted to inequality we needed to prove, so we are basically almost done with the proof.

Now, let's just recap everything. For any arbitrary $M \in \Bbb{R}$, we can choose a $N_\epsilon$ such that $n > N_\epsilon \rightarrow \frac{2M}{a} > \frac{M}{a_n}$. Also, we can choose an $N_O$ such that $n > N_O\rightarrow b_n > \frac{2M}{a}$. In order to use transitive property, we need both of these inequalities to be true. Thus, we can say if $n > \max(N_\epsilon, N_O)$, then both of the inequalities are true and by transitive property, $b_n > \frac{M}{a_n}\rightarrow a_nb_n > M$. Therefore, we have proven that for any $M\in \Bbb{R}$, there exists $N_M=\max(N_\epsilon, N_O)$ such that if $n > N_M$, then $a_nb_n > M$. Therefore, $\lim_{n\to \infty}a_nb_n=+\infty$.

Answer 2

To prove that $\displaystyle\lim_{n\rightarrow\infty} a_nb_n = \infty$ we need to show that for any

$$M\quad \exists \bar n \quad \forall n\ge \bar n\quad a_nb_n\ge M$$

therefore note that since $\displaystyle\lim_{n\rightarrow\infty} a_n= a > 0$

$$\exists \bar n_1 \quad \forall n\ge \bar n_1\quad a_n\ge \frac a2$$

and since $\displaystyle\lim_{n\rightarrow\infty} b_n=\infty$

$$\exists \bar n_2 \quad \forall n\ge \bar n_2\quad b_n\ge 2M$$

it suffices to chose $\bar n\ge \max\{\bar n_1,\bar n_2\}$.

Answer 3

Let $\epsilon={a\over 2}>0$ therefore there exists $N$ such that $$n>N\to |a_n-a|<\epsilon ={a\over 2}$$by multiplying both sides in $|b_n| $ we obtain$$|a_nb_n-b_na|<|b_n|{a\over 2}$$or equivalently$$-{a\over 2}|b_n|+b_na<a_nb_n<{a\over 2}|b_n|+b_na$$therefore $$\lim a_nb_n=\infty$$ by using Squeeze theorem.