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anonymous

  • 5 years ago

in need help with series...(1)/(n(n+2))

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  1. anonymous
    • 5 years ago
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    1/(n^2+2n)=1/n-1/(2*(n+2))

  2. anonymous
    • 5 years ago
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    Actually, \[ \frac{1}{n(n+2)} = \frac{1}{2n} - \frac{1}{2(n+2)} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2} \right) \] :(

  3. anonymous
    • 5 years ago
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    oh yes sorry

  4. anonymous
    • 5 years ago
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    so my book's examples show that 1/2[(1/(2n-1))-(1/(2n+1)) ] is the nth partial sum of a series and it equals 1/2[1-(1/(2n+1))].....this is what i don't understand because if i do that to 1/2[(1/n)-(1/(n+2))] making 1/n turn to a one then it does not give me the right answer ....i am missing something

  5. anonymous
    • 5 years ago
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    Write out the first 2/3 terms in the series (n = 1, 2, 3 ... more if you need to) Also write out the last 2/3 (n = n-2, n-1, n...) and see what cancels. There should be 4 fractions left, but it's better if you find out WHY they are left than just get told.

  6. anonymous
    • 5 years ago
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    ineton good job guy

  7. anonymous
    • 5 years ago
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    so I see that the example in the book makes sense now but i am still not seeing it for 1/2[(1/n)-(1/(n+2))].......is goes .5(1-1/3)+.5(1/2-1/4)+.5(1/3-1/5)+.5(1/4-1/6)..... .5[(1/n-3)-(1/n-1)]+.5[(1/n-2)-(1/n)]+.5[(1/n-1)-(1/n+1)....so i see that some of these fractions cancel but all like in the example....i am still not sure where to go from here

  8. anonymous
    • 5 years ago
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    oops but not* all

  9. anonymous
    • 5 years ago
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    yes never mind...i get it now....thanks so much!!!!!!

  10. anonymous
    • 5 years ago
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    Ah, OK, I was typing it out :( But you're welcome.

  11. anonymous
    • 5 years ago
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    well i have another problem...i really just having a problem with a limit....i keep trying to use L'Hopital's rule .....the series is \[\sum_{n=0}^{\infty}\] 1/(2^n)......so the nth partial sum is ((2^n)-1)/(2^(n-1)).....then i get the limit of the nth partial sum and that is where i am having trouble

  12. anonymous
    • 5 years ago
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    are you asking about the sum of\[\sum_{0}^{\infty}{1 \over 2^n}\]?

  13. anonymous
    • 5 years ago
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    well this can be written as: \[1+\sum_{n=1}^{\infty}{1 \over 2}({1 \over 2})^{n-1}\], which is a geometric series. the sum is \[1+{1/2 \over 1-1/2}=2\]

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