Re: Tea cups and elephants

On Nov 30, 2:46 am, William Hughes <wpihug...@xxxxxxxxxxx> wrote:
On Nov 29, 6:36 pm, matt271829-n...@xxxxxxxxxxx wrote:


<snip>


In the first case I have

lim n->oo (sum_i=1^n c/n) = c

Which hopefully is uncontroversial.

Indeed, Note that sum_i=1^n c/n = (n)*(c/n)

I'm having problems writing the
second case that represents the limit of summing the limits, which
leads to

0 + 0 + 0 + ... = 0

I want to write something like lim n->oo (sum_i=1^n (lim n->oo (c/n)))
= 0, but this doesn't seem right with the conflicting use of n.

Correct. you can only have one "lim n->oo"

I tried lim m->oo (sum_i=1^m (lim n->oo c/n)),

This is correct (note that this is (lim m->oo (m))*(lim n->oo c/n))

lim m->oo (sum_i=1^m (lim n->oo c/n)) seems to be

0 + 0 + 0 + ...

So, do we conclude that (lim m->oo (m))*(lim n->oo c/n) = lim m->oo
(sum_i=1^m (lim n->oo c/n)) = 0?


but having m and n
unrelated doesn't seem right either. Here we're evaluating the inner
limit first, and then summing, whereas I actually want the two limits
to be approached in tandem.

The two limits approach in tandem is lim n->oo (sum_i=1^n (c/n))


The formulation that I was searching for should have limits approached
in tandem, and yet result in something that we can explicitly see is
equal to 0 + 0 + 0 + ..., as is the case with lim m->oo (sum_i=1^m
(lim n->oo c/n)).

However, if there is only one possible way of evaluating the
expression with limits approached in tandem, and that is lim n->oo
(sum_i=1^n c/n), then presumably what I'm trying to do is impossible,
otherwise we would have lim n->oo (sum_i=1^n c/n) = 0 + 0 + 0 + ... =
0, which we know is wrong. Does that sound right?

I must admit I find it quite hard to see why lim n->oo (sum_i=1^n c/n)
is not equal to 0 + 0 + 0 + ... (To be clear, I am not suggesting that
c/n = 0 for any n; I'm talking about equal in the limit, in just the
same way that lim n->oo (c/n) = 0.)
.

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