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worm problem, PARADOX?

On March 9, 1997 John Mallinckrodt responded to the "paradox message" by
saying that 1+10^5 should not have been considered negligible. He wrote:

The "significance" of a term is always judged only in relation to
something else. No term is ever "insignificant" in absolute terms.
Fundamentally, this is where you erred.

I was neglecting the LST=1+10^5 term in the expression [L{n}-LST].
Thus I am comparing LST with L{n}, for a very large n (near the end of
the bug's race). Why do you say "in absolute terms"? Yes, I should have
used LST=1+10^5 instead of 10^5 but what difference does it make if LST
is to be ignored. How does my reasoning differ, for example, from showing
that the Riemann's sum is an approximation of an intergal (very large n)?

I have no doubt that there is an error in my reasonnig, somewhere, but
I still do not know where. The verbal part of what you are quoting could
be improved; I am not a mathematician. What I would like to see is the
derivation of the correct finite value of n from the iterative formula
for x{n}. The "paradox was discovered" when I was unable to do this. The
original motion algorithm was described iteratively and it would be nice
to find the answer without introducing the harmonic series.

Once again I would like to thank you, John, for clearly explaining the
correct "non-iterative" solution.

********************************************************************
HERE IS JOHN'S WHOLE MESSSAGE (HE IS RESPONDING TO WHAT IS QUOTED WITH >):

At time t=n the tractor's location (along the rigid x axis) is

L{n}=10^5+10^5*n=10^5*(n+1) (A)

The location of the bug, at t=n, is given by iterative formula

x{n}=[x{n-1}+1]*(1+1/n) (B)

That formula yields exactly the same values as the non-iterative formula
which contains a sum of 1/k terms (Uri's notation). As everybody else, I
say that the bug will catch the tractor when L{n}=x{n}. At that moment

x{n}-x{n-1}=LST=10^5 (C)

where LST is the length of the last step.

No. I assume that you mean for LST to be *defined* to be x{n} - x{n-1}.
If so, then LST is approximately equal to but slightly greater than 10^5.
In fact LST is exactly equal to 100001. There is a world of difference
between approximately equal and equal. You needn't exercise any particular
care in applying an equality; you must exercise extreme care in applying
an approximation.

(the last speed of the bug is 10^5 cm/s, the same as the speed of the
tractor).

I don't know how to interpret this in any precise way, but it doesn't
appear to be critical to your argument so I'll simply let it slide.

I know, from the sollutions of Chip, Uri or John, that x{n} and x{n-1}
are gigantic in comparison with 10^5. What is one kilometer (10^5 cm)
in comparison with zilions of trip around the universe? Therefore I can
assume, while writing L{n}=x{n}, that x{n}=x{n-1}.

No; x{n} is approximately equal to but slightly larger than x{n-1}. It is
in fact larger by 100001. Again, there is a world of difference between
approximately equal and equal.

Once I do this I conclude that 1/n --> 0 and n --> infinity, as shown
below. What is wrong?

L{n}=[x{n-1}+1]*(1+1/n) or (D)
Yes
L{n}=[L{n}-LST]*(1+1/n) or (E)

No; L{n} is approximately equal to but slightly larger than
[L{n}-LST]*(1+1/n). In fact, L{n}=[L{n}-LST]*(1+1/n) + (1+1/n)

L{n}=[L{n}-0]*(1+1/n) (F)

No; L{n} is approximately equal to but slightly less than [L{n}-0]*(1+1/n).
In fact, L{n}=[L{n}-0]*(1+1/n) - 10^5*(1+1/n). If you subtract the enormous
value L{n} from both sides and then simplify this equation you get
L{n} = 10^5*(n+1) as we already knew. No paradox here.

I know that n=10^43429.443 and n--> infinity are practically identical.

(This might qualify as the most incorrect thing you've said.)

But conceptually they are very different. How can a neglection of an
insignificant term, LST, change a finite solution of a problem into
infinity? Where did I err?

The "significance" of a term is always judged only in relation to
something else. No term is ever "insignificant" in absolute terms.
Fundamentally, this is where you erred.
John