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anonymous
 5 years ago
in need help with series...(1)/(n(n+2))
anonymous
 5 years ago
in need help with series...(1)/(n(n+2))

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/(n^2+2n)=1/n1/(2*(n+2))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, \[ \frac{1}{n(n+2)} = \frac{1}{2n}  \frac{1}{2(n+2)} = \frac{1}{2} \left(\frac{1}{n}  \frac{1}{n+2} \right) \] :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so my book's examples show that 1/2[(1/(2n1))(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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Write 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 = n2, n1, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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(11/3)+.5(1/21/4)+.5(1/31/5)+.5(1/41/6)..... .5[(1/n3)(1/n1)]+.5[(1/n2)(1/n)]+.5[(1/n1)(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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes never mind...i get it now....thanks so much!!!!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, OK, I was typing it out :( But you're welcome.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well 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^(n1)).....then i get the limit of the nth partial sum and that is where i am having trouble

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you asking about the sum of\[\sum_{0}^{\infty}{1 \over 2^n}\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well this can be written as: \[1+\sum_{n=1}^{\infty}{1 \over 2}({1 \over 2})^{n1}\], which is a geometric series. the sum is \[1+{1/2 \over 11/2}=2\]
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