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Which of the following is Cauchy's sequence ?a)(-1)^n+1/n b)(-1)^n.n c) n+1...
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-1+1, 1+1/2, -1+1/3, 1+1/4, -1+1/5, 1+1/6, -1+1/7, 1+1/8
every second number is getting close to 1, and the other numbers are getting close to -1.
The 1/n part of the expression is getting close to 0, but the (-1)^n part of the expression is bouncing between -1 and 1
Thus this sequence is not Cauchy.
Definition of Cauchy: for every epsilon, there is some N such that, if n,m>N then An and Am are within epsilon of each other.
Pick epsilon = 1
No matter what N you pick, if you look at An and A(n+1), they will be about 2 apart: one will be close to 1 and one will be close to -1, and the difference between these will be more than 1
-1, 2, -3, 4, -5, 6, -7, 8, ...
These numbers are unbounded, so they are not convergent. In the real numbers, a Cauchy sequence is the same as a convergent sequence. This sequence is unbounded, so it is not Cauchy.
c) 1+1, 2+1/2, 3+1/3, 4+1/4, 5+1/5, ...
Again, these numbers are unbounded, so they are not Cauchy
-1, 1/2, -1/3, 1/4, -1/5, 1/6, ...
These numbers are converging to 0. Again, in the real numbers, Cauchy is convergent, so this is a Cauchy sequence. Or, to see this rigorously:
Pick some `epsilon` . We want to find N such that, for n,m>N, `|A_n - A_m|<epsilon`
`|A_n - A_m| <= |A_n| + |A_m| = 1/n + 1/m < 2/m`
(if we assume m `<=` n ... one of them has to be less than or equal to the other)
Thus, we want `2/m < epsilon`
So: Pick N so that `N>2/epsilon`
Then, if m, n>N we have `1/m < epsilon / 2` and `1/n<epsilon /2`
`|A_n-A_m| <= 1/m+1/n < epsilon/2 + epsilon/2 = epsilon`
Thus, sequence 4 is Cauchy
Posted by mlehuzzah on December 27, 2012 at 10:06 PM (Answer #1)
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