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Prove that a_n tends to 0

Alexmahone

Active member
Jan 26, 2012
268
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 $\epsilon<1$
 
Last edited:

ThePerfectHacker

Well-known member
Jan 26, 2012
236
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).