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 $\displaystyle n$.
$\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}
$
Of course you are right, I don't mean what I say at all do I .
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