pahio,
The limit of 2n/(n+1) as positive n goes towards infinity is 2.
0/1, 2/2, 4/3, 6/4, ... 1998/1000.
[The number 999 was a very astute choice for n.]
1.98, 1.998, 1.9998, 1.99998, gets closer and closer to 2.
Now 1.98 + .02 = 2; 1.998 + .002 = 2; 1.9998 + .0002 = 2; ... 1.9999...98 + .0000...2 = 2.
If everybody is willing I'd like to call .0000...2 the error, with the symbol lower case epsilon.
For negative n, the limit of 2n/(n-1) as negative n goes to negative infinity is -2.
Re: The sequence 1, -1, 1, -1, ... +/- 1.
The series 1 + (-1) + 1 + (-1) + 1 + (-1)... +/-1 = 1, 0, 1, 0, 1, ...; has wavy behavior, and may be hidden in other sequences.
We could also write 0.9, 0.99, 0.999, ... 0.9999...9 as (1-(10^{-n})).
Unfortunately, marie makes lots of mistakes, and she can't get through a page of calculations without messing up the signs. Worse yet, she'd think of a possibility like (1-(10^{-(-n)}).
Thank you for all the fruitful ideas.
- marie |
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