27 Oct 14
Originally posted by adam warlockNo, you've already shown the sum is convergent by summing it. The ratio test works by comparing a series with the series 1 + x + x² + x³ + ... whose sum is known to be finite, so using the ratio test to prove the series with x = 1/2 is finite is begging the question.
This deduction assumes that the sum is convergent (in fact I think that it assumes that the sum is absolutely convergent (which it is) since you're making term by term matching) so in interest of completeness one must prove that the sum indeed is (absolutely) convergent.
For the series S, we have the self similarity relation:
S = 1 + x + x² + x³ + ...
= 1 + x (1 + x + x² + ...)
= 1 + x S
so
S = 1/(1 - x)
and the series is convergent for all |x| < 1.
It is divergent if x = 1 and for all x > 1, Since the nth partial sum >= n.
It is conditionally convergent for all x <= -1 (and sums to 1/2 for x = -1).
27 Oct 14
3 edits
Originally posted by DeepThoughtNo you haven't. By summing it you're assuming that it is convergent, and if S is convergent then it has the value you found.
No, you've already shown the sum is convergent by summing it.
-1+1-1+1-1+...=0 by the same logic.
Edit: Another example of the failure of that logic: let "a>0".
S = …1/a² + 1/a + 1 + a + a²…
aS = …1/a + 1 + a + a² + a³… = S
aS = S
S=0
27 Oct 14
1 edit
Originally posted by adam warlockThe series is the index case as far as convergence of series go. We'd discussed it on page 2 for the special case that x = 1/2. For the general case the nth partial sum S(n,x) = (1 - x^(N+1))/ (1-x) < 1/(1 - x) for all N provided that 0 < x < 1. So since the sequence of partial sums is bounded from above the series is convergent and sums to the limit of the sequence.
No you haven't. By summing it you're assuming that it is convergent, and if S is convergent then it has the value you found.
-1+1-1+1-1+...=0 by the same logic.
Edit: Another example of the failure of that logic: let "a>0".
S = …1/a² + 1/a + 1 + a + a²…
aS = …1/a + 1 + a + a² + a³… = S
aS = S
S=0
27 Oct 14
2 edits
The post that was quoted here has been removedaS = S
Subtract S from both sides:
aS – S = 0
factorize:
S(a – 1) = 0
Divide (a – 1) from both sides:
S = 0/(a – 1)
and providing the variable a doesn't equal 1 (else the fraction is nonsense because you cannot have the denominator equal to zero ) :
S = 0
Note though that, if variable a equals 1, S may not equal zero.
28 Oct 14
Originally posted by Great King RatI am defining S ... you cannot set it to any number.
Why is 2S - S = 1???
If I fill in S=3, then 2S - S = 2*3 - 3 = 3.
You filled in S= 1/2 + 1/4 + 1/8 + .....
Therefore 2S - S = 1/2 + 1/4 + 1/8 + .....
Right??
S is the sum of the sequence ... it's what we are trying to find.
We subtract S from 2S to leave 1.
(Yes S=1 and S= 1/2 + 1/4 + ...)