Thread: 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.

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:



for large , and as the RHS goes to infinity does as well.

Now all you have to do is write out what means in full and rearrange it to show that
differers form by arbitrarily small amounts for sufficiently large .

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

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 .

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