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

1. anonymous

1/(n^2+2n)=1/n-1/(2*(n+2))

2. anonymous

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

oh yes sorry

4. anonymous

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

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

ineton good job guy

7. anonymous

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

oops but not* all

9. anonymous

yes never mind...i get it now....thanks so much!!!!!!

10. anonymous

Ah, OK, I was typing it out :( But you're welcome.

11. anonymous

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

are you asking about the sum of$\sum_{0}^{\infty}{1 \over 2^n}$?

13. anonymous

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 $1+{1/2 \over 1-1/2}=2$