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blackstreet23

  • one year ago

I am trying to do an alternating series test problem but i forgot how to prove if the series is decreasing. I remember they were several tests for that, but i forgot which where. Please help!

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  1. freckles
    • one year ago
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    Can you show me your series?

  2. freckles
    • one year ago
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    Sometimes it is pretty easy with derivative. Sometimes it is easy with algebra. like we can show \[a_{n+1}<a_n \text{ for }n \text{ using algebra } \\ \text{ or show } f' \text{ is negative for all } x\]

  3. freckles
    • one year ago
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    n not x

  4. blackstreet23
    • one year ago
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    could you tell me all the tests? I am trying to review my calc 2 to be ready for calc 3

  5. freckles
    • one year ago
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    Those are the only two I can recall.

  6. SolomonZelman
    • one year ago
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    In alternating series, if the absolute value of the terms decreases then the series will converge.

  7. blackstreet23
    • one year ago
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    1 Attachment
  8. SolomonZelman
    • one year ago
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    So, I will denote the sums with \(\rm S_1,~S_2...\) |dw:1441594022749:dw|

  9. blackstreet23
    • one year ago
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    that is what i have so far

  10. SolomonZelman
    • one year ago
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    Can you post it in here, more clearly.....

  11. freckles
    • one year ago
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    Is the attachment bringing up a new question?

  12. blackstreet23
    • one year ago
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    no just showing my work so far

  13. freckles
    • one year ago
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    \[k+1>k \text{ so } \frac{1}{k+1}<\frac{1}{k}\]

  14. freckles
    • one year ago
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    or. ... \[f(k)=\frac{1}{k+1} \\ f'(k)=\frac{-1}{(k+1)^2} \\ \frac{-1}{(k+1)^2}<0 \text{ for all } k \text{ except } k=-1 \\ \text{ which we aren't concerned about }\] either one of these ways shows a_k is a decreasing sequence

  15. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sum_{ k=0 }^{ \infty } \frac{x^k}{k+1}}\) \(\large\color{black}{ \displaystyle \lim_{ k\rightarrow \infty }\frac{x^{k+1}(k+1)}{(k+2)x^k}}\) \(\large\color{black}{ \displaystyle x\lim_{ k\rightarrow \infty }\frac{k+1}{k+2}}\) \(\large\color{black}{ \displaystyle|x |<1 }\)

  16. SolomonZelman
    • one year ago
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    and it is actually good, because even if it was just x\(^k\) without dividing by k+1, then it converges for |x|<1. And when you divide by k+1, then it is even smaller. (And x=-1 is included in convergence, because it is alternating harmonic series when x=-1)

  17. SolomonZelman
    • one year ago
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    So it should really be: \(-1 \le x<1\)

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