Name: Dane
Who is asking: Other
Level: All
Question:
Something I noticed fooling around with a calculator about 30 years
ago.
Considering e = 2.718281828459045....
Using Window's Calculator you will find
1.111 = 2.8531167...
1.01101 = 2.731861...
1.0011001 = 2.71964085...
1.000110001 = 2.71841774...
1.00001100001 = 2.7182954...
1.00000110000011 = 2.178231875...
1.000000110000001 = 2.178219643...
There apears to be a pattern. My conjecture is:
1.'infinite number of zeros'11'infinite number of zeros'1 = e.
I have no idea if this is true and no idea on how to go about proving
my conjecture one way or another. If you extend the pattern
using a calculator you will find the answer eventually starts diverging
from e. I think this
is due
to calculations
being lost
in the floating point round off fuzz inherent in all computers.
Any insight you may have would be enlightment.
Thank You
Dane
Hi Dane,
That's a really nice observation.
One of the facts about the number e is that the sequence
(1+1)1, (1+ 1/2)2, (1+ 1/3)3, (1+ 1/4)4,...
approaches e. That is
the limit, as n approaches infinity, of (1 + 1/n)n is e
Some authors take this fact as the definition of e and some authors define e some other way and then prove this fact as a theorem.
Your sequence is (almost) every tenth term of the sequence (1 + 1/n)n. Your sequence is (1 + 1/n)n+1 where n is 10, 100, 1000, ...
Since the sequence (1 + 1/n)n approaches
e, the subsequence of every tenth term approaches e also. You multiply
the terms of this sequence by the sequence (1+ 1/n).
But the limit of the sequence (1+ 1/n) is
1 and
hence the limit of your sequence is 1
e = e.
Penny