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July 3rd, 2017, 07:25 AM   #1
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Limit consequence

Is consequence true ?
If $\displaystyle \lim_{n\rightarrow \infty} \frac{a_n}{b_n}=0 \; \; \Rightarrow \;
\; b_n\geq a_n$
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July 3rd, 2017, 08:17 AM   #2
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No. $a_n = 0, b_n = -1$

If you meant to have both positive
\begin{align*}
a_n &= n \\
b_n &= \begin{cases}
n^2 & (n \gt N) \\
1 & (n \le N)
\end{cases}
\end{align*}

The answer to the question you intended to ask (i.e. for sufficiently large $n$) is yes. Think about $\delta-\epsilon$ definition.

$\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = 0$ means that for every $\epsilon \gt 0$ there exists an $N \in \mathbb N$ such that $\frac{a_n}{b_n} \lt \epsilon$ for all $n \gt N$.

Pick $ \epsilon = 1$ and you have your result.

Last edited by v8archie; July 3rd, 2017 at 08:40 AM.
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July 3rd, 2017, 08:39 AM   #3
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i forgot to check $\displaystyle a_n,b_n >0$
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July 3rd, 2017, 08:40 AM   #4
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See update.
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July 4th, 2017, 06:40 AM   #5
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Quote:
Originally Posted by idontknow View Post
Is consequence true ?
If $\displaystyle \lim_{n\rightarrow \infty} \frac{a_n}{b_n}=0 \; \; \Rightarrow \;
\; b_n\geq a_n$
No.

$\displaystyle \left | \frac{a_{n}}{b_{n}} \right |\leq \epsilon\\
|a_{n}| \leq|b_n|$,
n>N

Ref: v8archie post#2.

Last edited by zylo; July 4th, 2017 at 06:51 AM. Reason: add n>N
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