Re: Tea cups and elephants

On Nov 30, 2:14 pm, Randy Poe <poespam-t...@xxxxxxxxx> wrote:
On Nov 30, 9:05 am, matt271829-n...@xxxxxxxxxxx wrote:





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.)

Because as far as I can tell you're still thinking of
"in the limit" as an actual evaluation of an expression
with variables "set to infinity".

I don't think I am ... see later comment.


You have two different sequences. One of them is
{0,0,0,... }, i.e. 0 for every finite n. The other is
{1,1,1,... }, i.e. 1 for every finite n. Does it
really seem that {0,0,0,...} should eventually reach
1 "in the limit"?

No.


I think the problem may also be in insisting that
the sum "in the limit" can be written as something
involving plus signs, even when there are an
infinite number of terms "in the limit".

By "0 + 0 + 0 + ..." I mean the limit as n->oo of the sum of n zeroes.
I am assuming that this is a well-defined notion, and that it equals
zero.

What is
actually happening is that there is, as I wrote,
a sequence of numbers. When you consider the numbers,
the value of each finite term, I hope your intuition
doesn't still suggest that {0,0,0,...} should
"eventually" reach 1.

It seems clear to me that 0 + 0 + 0 + ... (as defined above) must
equal zero. On the other hand, it also seems clear that sum_i=1^n c/n
approaches 0 + 0 + 0 + ... as a limiting case, so that when we find
the limit as n->oo we actually *do* get 0 + 0 + 0 + ... I presume that
this is wrong because lim n->oo (sum_i=1^n c/n) is clearly equal to c.
However, I can't quite grasp why we get the wrong answer. Do you see
what I'm getting at?

Note that I am not assuming that the terms actually equal zero for any
finite n, any more than I believe lim n->oo (c/n) = 0 implies c/n = 0
for some n. In neither case is anything evaluated with n "set to
infinity".
.

Relevant Pages

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