# A limit

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• Nov 1st 2006, 07:33 PM
JimmyT
A limit
Need serious help on a proof:
Show that if lim(an/n) = L where L>0, then lim an (n approaches inf) = inf.

Thanks guys.
• Nov 1st 2006, 09:25 PM
CaptainBlack
Quote:

Originally Posted by JimmyT
Need serious help on a proof:
Show that if lim(an/n) = L where L>0, then lim an (n approaches inf) = inf.

Thanks guys.

Informally this is because if:

$\displaystyle \lim_{n \to \infty}a_n/n = L\ >0\ \ \ \dots\ (1)$

for large $\displaystyle n,\ a_n \sim nL$, and as the RHS goes to infinity $\displaystyle a_n$ does as well.

Now all you have to do is write out what $\displaystyle (1)$ means in full and rearrange it to show that $\displaystyle a_n$
differers form $\displaystyle nL$ by arbitrarily small amounts for sufficiently large $\displaystyle n$.
• Nov 2nd 2006, 11:26 AM
JimmyT
I don't understand what you mean when you say "rearrange" it by "arbitrarily small amounts."
• Nov 2nd 2006, 12:15 PM
Plato
$\displaystyle \begin{array}{l} L > 0\quad \Rightarrow \quad \left( {\exists N} \right)\left[ {n \ge N \Rightarrow \left| {\frac{{a_n }}{n} - L} \right| < \frac{L}{2}} \right] \\ \Rightarrow \quad \frac{L}{2} < \frac{{a_n }}{n} \\ \Rightarrow \quad n\frac{L}{2} < a_n \\ \end{array}$
• Nov 2nd 2006, 01:34 PM
CaptainBlack
Quote:

Originally Posted by JimmyT
I don't understand what you mean when you say "rearrange" it by "arbitrarily small amounts."

Of course you are right, I don't mean what I say at all do I:mad: .

What I do mean is that for every epsilon>0, there exists an N such that
for all n>N:

-epsilon< (a_n)/n-L <epsilon

so:

nL-n epsilon< a_n < nL + n epsilon,

or:

n(L-epsilon) < a_n < n(L+epsilon)

so if we choose a small enough epsilon, as n-> infnty a_n is trapped between
two sequences both of which go to infinity.

RonL