MHB Does a_n Converge to Zero Given b_n Approaches Zero and a_n/b_n Approaches L?

[alexmahone]

Prove: if $\frac{a_n}{b_n}\to L$, $b_n\neq 0$ for any $n$, and $b_n\to 0$, then $a_n\to 0$. ($L$ represents a finite number, not $\infty$.)

My working:

Given $\epsilon>0$,

$\left|\frac{a_n}{b_n}-L\right|<\epsilon$ and $|b_n|<\epsilon$

How do I proceed?

---------- Post added at 11:43 AM ---------- Previous post was at 10:55 AM ----------

Let me give it a shot.

$\left|\frac{a_n}{b_n}\right|=\left|\frac{a_n}{b_n}-L+L\right|\le\left|\frac{a_n}{b_n}-L\right|+|L|<\epsilon+|L|$

$|a_n|<|b_n|(\epsilon+|L|)<\epsilon(\epsilon+|L|)<\epsilon(1+|L|)$

since we may assume $\epsilon<1$

 

[ThePerfectHacker]

Hello,

Let us take \( \varepsilon = 1 \) there is a large enough integer \( N \) such that if \( n\geq N \) then, \( \big| \frac{a_n}{b_n} - L \big | < \varepsilon = 1 \). This means that (opening up the absolute value),

$$ -1 < \frac{a_n}{b_n} - L < 1 \implies L-1 < \frac{a_n}{b_n} < L+1 $$

Now multiply both sides by \( b_n \), but be careful! If \( b_n \) happens to be negative the inequality gets reversed. Therefore, this means that, \( (L-1)b_n < a_n < (L+1)b_n \) if \(b_n > 0 \) and \( n\geq N \). And that \( (L+1)b_n < a_n < (L-1)b_n \) if \( b_n < 0 \) and \( n\geq N \).

Now as \( b_n \to 0\) by the Squeeze Theorem the sequence \( a_n \to 0\) also (notice that whether \( b_n \) is positive or negative does not matter as in both cases these sequences are going to zero anyway).

 

[Greg Bernhardt]

Here's one way to proceed:

First, we can rewrite the first inequality as:

$$-\epsilon<\frac{a_n}{b_n}-L<\epsilon$$

Multiplying by $b_n$ (note that $b_n$ is positive since we are given $b_n\neq 0$) gives:

$$-b_n\epsilon<a_n-Lb_n<b_n\epsilon$$

Since we are also given that $b_n\to 0$, we know that for any $\delta>0$, there exists an $N$ such that for all $n>N$, $|b_n|<\delta$. So, let's choose $\delta = \frac{\epsilon}{2|L|}$ (note that we can make this choice since we are given that $L$ is finite, not $\infty$).

Then, for all $n>N$, we have $|b_n|<\frac{\epsilon}{2|L|}$, which implies that $b_n\epsilon<\frac{\epsilon^2}{2|L|}$. So, we have:

$$-b_n\epsilon<a_n-Lb_n<\frac{\epsilon^2}{2|L|}$$

Now, we can add $Lb_n$ to all parts of the inequality to get:

$$L-\frac{\epsilon^2}{2|L|}<a_n<L+\frac{\epsilon^2}{2|L|}$$

Since $\epsilon^2$ is positive, we can rewrite this as:

$$L-\frac{\epsilon^2}{2|L|}<a_n<L+\frac{\epsilon^2}{2|L|}$$

Now, since $|L|$ is finite (we are given that $L$ is finite, not $\infty$), we can choose $\epsilon$ small enough so that $\frac{\epsilon^2}{2|L|}<\frac{\epsilon}{2}$. This gives us:

$$L-\frac{\epsilon}{2}<a_n<L+\frac{\epsilon}{2}$$

But this is exactly the definition of $a_n\to L$. So, for any $\epsilon>0$, we can find an $N$ such that for all $n>N$, $|a_n-L|<\epsilon$, which means that $a_n\to L$.