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anonymous

  • 5 years ago

why does the series {1/2^n} from n=1 to infinity converge?

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  1. anonymous
    • 5 years ago
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    By the ratio test, letting\[a_n=\frac{1}{2^n}\]then\[\lim_{n \rightarrow \infty}\left| \frac{a_{n+1}}{a_n} \right|=\lim_{n \rightarrow \infty}\left| \frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}} \right|=\lim_{n \rightarrow \infty}\frac{2^n}{2^{n+1}}=\lim_{n \rightarrow \infty}\frac{1}{2}=\frac{1}{2}<1\]

  2. anonymous
    • 5 years ago
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    according to a theorem that all geometric series are convergent.

  3. anonymous
    • 5 years ago
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    If the magnitude of the ratio of successive terms is less than 1, it is convergent.

  4. anonymous
    • 5 years ago
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    ok , why does it not ->0 and diverge by nth term test? and thanks!

  5. anonymous
    • 5 years ago
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    The nth term test says, if in the limit, your terms DON'T go to zero, then the series WON'T converge. That's all it says.

  6. anonymous
    • 5 years ago
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    And 1/2^n does go to zero as n approaches infinity.

  7. anonymous
    • 5 years ago
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    awesome, and i could use the a/1-r with a=1/2 and r=1/2^2 to find the same thing as the ratio test?

  8. anonymous
    • 5 years ago
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    If it's a geometric series with common ration less than 1, you can use the formula, though you have to show first that you do in fact have a geometric series (even though it's obvious here).

  9. anonymous
    • 5 years ago
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    *ratio

  10. anonymous
    • 5 years ago
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    ok, lastly, what would the ratio be in that case?

  11. anonymous
    • 5 years ago
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    1/2

  12. anonymous
    • 5 years ago
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    Divide the (n+1)th term by the nth term to see.

  13. anonymous
    • 5 years ago
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    ok so the ratio only has to get you from s1 to s2 and not any further?

  14. anonymous
    • 5 years ago
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    What's s1 and s2?

  15. anonymous
    • 5 years ago
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    The terms in your sequence?

  16. anonymous
    • 5 years ago
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    partial sum 1 ( which would be the index) and partial sum 2 (which should equal index*ratio) ?

  17. anonymous
    • 5 years ago
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    The partial sum for a geometric series is something different to the limiting sum that you use (i.e. your formula). The ratio of two successive partial sums will not, in general, be the common ratio.

  18. anonymous
    • 5 years ago
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    The terms you test are the terms in the *sequence* that make up the sum.

  19. anonymous
    • 5 years ago
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    ok ok. thanks so much, cleared up a lot of questions i was wasting a lot of time trying to figure out on my own. =)

  20. anonymous
    • 5 years ago
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    You're welcome :)

  21. anonymous
    • 5 years ago
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    one more, why is \[2^{n}/ 2^{n+1} = 1/2 \] ?

  22. anonymous
    • 5 years ago
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    Consider \[\frac{2^4}{2^5}\]The definition of that notation is such that\[2^4=2 \times 2 \times 2 \times 2 \]and\[2^5=2 \times 2 \times 2 \times 2 \times 2 \]so\[\frac{2^4}{2^5}=\frac{2 \times 2 \times 2 \times 2 }{2 \times 2 \times 2 \times 2 \times 2 }=\frac{1}{2}\]

  23. anonymous
    • 5 years ago
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    If you have, in general,\[\frac{a^n}{a^m}\]then it's the case that\[\frac{a^n}{a^m}=a^{n-m}\]

  24. anonymous
    • 5 years ago
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    In your case,\[\frac{2^n}{2^{n+1}}=2^{n-(n+1)}=2^{-1}=\frac{1}{2}\]since\[a^{-p}=\frac{1}{a^p}\]

  25. anonymous
    • 5 years ago
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    its always the exponent algebra =/

  26. anonymous
    • 5 years ago
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    Yeah, you need to come to grips with that before you embark on sequences and series.

  27. anonymous
    • 5 years ago
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    def. thanks again =)

  28. anonymous
    • 5 years ago
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    np

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