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Convergence of alternating series

  • Thread starter vikkisut88
  • Start date
1. Homework Statement
Let s be the sum of the alternating series [tex]\sum[/tex](from n=1 to [tex]\infty[/tex])(-1)n+1an with n-th partial sum sn. Show that |s - sn| [tex]\leq[/tex]an+1



2. Homework Equations
I know about Cauchy sequences, the Ratio test, the Root test


3. The Attempt at a Solution
I really have no idea where i'm meant to start with this question. I didn't know whether I thne have to take more partial sums i.e. partial sums of the n-th partial sum and work with it from there, or something else. I would be really grateful if someone could help me out.
 

CompuChip

Science Advisor
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I don't think your statement is true as you gave it.
For example, let
[tex]\begin{cases}
a_0 & {} = 1 \\
a_1 & {} = 0 \\
a_n & {} = 2^{-n} \qquad (n > 1)
\end{cases}[/tex]

Then s = 3/2, |s - s0| = |5/6 - 1| = |-1/6| = 1/6 which is not smaller than a1 = 0.

[edit]I had s = 5/6, I think it should be 3/2. Doesn't affect my argument though.[/edit]
 
Last edited:
sorry i don't really understand that - how did you work out that s was 5/6? And did you just choose random values for a0, a1 and an? I have rechecked my homework question and that is exactly what it said!
 

Dick

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CompuChip just gave you a counterexample. To prove your conclusion, you also need to assume that a_n is a nonnegative sequence decreasing to zero monotonically. That might be what you are missing.
 
okay, but i still don't understand how i'm meant to show the result, sorry. This question has got me completely flummoxed.
 

Dick

Science Advisor
Homework Helper
26,258
618
Start by looking at the difference of two partial sums |s_m-s_n| for m<n. Try to show that's less than or equal to a_m+1. Write down the terms making up the difference and regroup them. This would show the sequence is Cauchy, right?
 

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