I have a problem that states the following:
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Form the series: a - (1/2)b + (1/3)a - (1/4)b + (1/5)a - (1/6) b + ...
a) Express this series in sigma notation.
b) For what positive values of a and b is this series absolutely convergent? conditionally convergent?
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Now, I'm having trouble figuring out how to approach this problem. Would it be valid to consider a single "term" of the sequence as having both an a and b part?
aka. write it as the sum from k to infinity of: a/(2k - 1) + b/(2k).
This screws up things when I try to consider absolute convergence, though, right (due to truths like the triangle inequality)? However, the only other way I see to write the sequence in sigma notation is with manipulations of strange powers (such as alternating 1 and -1 with some clever dividing, or with sines and cosines), which is disgusting and not intuitive.
As for the second part, I just draw a blank. If a = b, then it's an alternating harmonic series multiplied by a constant, which is conditionally convergent. What about a > b or b > a?
Thanks for any input you may have. :)
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Form the series: a - (1/2)b + (1/3)a - (1/4)b + (1/5)a - (1/6) b + ...
a) Express this series in sigma notation.
b) For what positive values of a and b is this series absolutely convergent? conditionally convergent?
--------------------------------------…
Now, I'm having trouble figuring out how to approach this problem. Would it be valid to consider a single "term" of the sequence as having both an a and b part?
aka. write it as the sum from k to infinity of: a/(2k - 1) + b/(2k).
This screws up things when I try to consider absolute convergence, though, right (due to truths like the triangle inequality)? However, the only other way I see to write the sequence in sigma notation is with manipulations of strange powers (such as alternating 1 and -1 with some clever dividing, or with sines and cosines), which is disgusting and not intuitive.
As for the second part, I just draw a blank. If a = b, then it's an alternating harmonic series multiplied by a constant, which is conditionally convergent. What about a > b or b > a?
Thanks for any input you may have. :)
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a) Σ(k = 1 to ∞) [a/(2k-1) - b/(2k)].
b) Assuming that a, b > 0:
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First, we determine when this series actually does converge.
This we accomplish with the Alternating Series Test.
(i) Plainly these terms converge to 0.
(ii) We need these terms (in absolute value) to form an eventually decreasing sequence:
For sufficiently large k, we need a/(2k-1) > b/(2k) and b/(2k) > a/(2k+1)
==> 2ak > (2k-1)b and (2k+1)b > 2ak
==> 2(a-b)k > -b and b > 2(a-b)k
These are both true iff a = b.
Hence, the series converges iff a = b.
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This makes the rest of the problem easy, since we can write the series as
Σ(k = 1 to ∞) a * (-1)^(k-1)/k.
For absolute/conditional convergence, we need to determine whether Σ(k = 1 to ∞) a/k converges.
This is not the case, because this is a multiple of the harmonic series, which is known to diverge.
Hence, the convergence is only conditional.
I hope this helps!
b) Assuming that a, b > 0:
-------
First, we determine when this series actually does converge.
This we accomplish with the Alternating Series Test.
(i) Plainly these terms converge to 0.
(ii) We need these terms (in absolute value) to form an eventually decreasing sequence:
For sufficiently large k, we need a/(2k-1) > b/(2k) and b/(2k) > a/(2k+1)
==> 2ak > (2k-1)b and (2k+1)b > 2ak
==> 2(a-b)k > -b and b > 2(a-b)k
These are both true iff a = b.
Hence, the series converges iff a = b.
-----------
This makes the rest of the problem easy, since we can write the series as
Σ(k = 1 to ∞) a * (-1)^(k-1)/k.
For absolute/conditional convergence, we need to determine whether Σ(k = 1 to ∞) a/k converges.
This is not the case, because this is a multiple of the harmonic series, which is known to diverge.
Hence, the convergence is only conditional.
I hope this helps!