i am looking for an explicit formula for the sequence given by the following numbers:

0, 1/2, 1, 2/3, 1/3, 0, 1/4, 2/4, 3/4, 1, 4/5, 3/5, 2/5, 1/5, 0, 1/6, etc..

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- Jul 4th 2009, 10:24 AMkkoutsothodorosformula for a sequence
i am looking for an explicit formula for the sequence given by the following numbers:

0, 1/2, 1, 2/3, 1/3, 0, 1/4, 2/4, 3/4, 1, 4/5, 3/5, 2/5, 1/5, 0, 1/6, etc.. - Jul 5th 2009, 02:50 AMmalaygoel
I could get upto this:

nth term,$\displaystyle t_n=\frac{d-m}{d}$

where

$\displaystyle d=\left\lceil\frac{\sqrt{8n+1}-1}{2}\right\rceil$

$\displaystyle m=(-1)^d \left( \frac {d(d+(-1)^d)}{2}-n \right)$

someone, please silmpify - Jul 5th 2009, 06:46 AMkkoutsothodoros
$\displaystyle d_{n}=\left\lfloor\sqrt{2n}+\frac{1}{2}\right\rflo or$

- Jul 5th 2009, 07:12 AMkkoutsothodoros
we get the general term of the sequence to be:

$\displaystyle a_{n}$=$\displaystyle \sum_{n=1}^{n}\frac{(-1)^{d_{n}}}{d_{n}}\$

i figured this out yesterday but didn't have the time to learn LaTex. And I had Mr. Fantastic sending me demerits all day long. maybe he could have spent more time helping with this as opposed to sitting in front of his computer sending out demerits!! - Jul 5th 2009, 07:21 AMkkoutsothodoros
i will probably get thrown off the site now!! well it was fun learning a little bit of latex!

- Jul 5th 2009, 07:44 AMkkoutsothodoros
oh yes, one more thing,

$\displaystyle a_{0}$=0. - Jul 5th 2009, 07:55 AMmalaygoel
- Jul 5th 2009, 08:22 AMkkoutsothodoros
it does seem a little unclear. sorry, i am just learning how to use latex.

let $\displaystyle a_{n}$ = $\displaystyle \frac{(-1)^{d_{n}}}{d_{n}}\$ for n >= 2, $\displaystyle a_{1}$ = 0.

let $\displaystyle s_{n}\$ = $\displaystyle \sum_{n=1}^{n}a_{n}$.

you should get

$\displaystyle s_{1}$ = 0

$\displaystyle s_{2}$ = .5

$\displaystyle s_{3}$ = 1

$\displaystyle s_{4}$ = 2/3

etc. - Jul 5th 2009, 08:24 AMkkoutsothodoros
so $\displaystyle s_{n}$ is the general term for the sequence 0, 1/2, 2/2, 2/3, 1/3, 0/3, 1/4, 2/4, 3/4, 4/4, 4/5, 3/5, etc

- Jul 5th 2009, 08:35 AMkkoutsothodoros
the funny thing is this sequence is just used as a counterexample in a cauchy sequence problem i was looking at. it wasn't that important to get the general term formula. it just bugged me so much that i couldn't let it go. but i guess it's useful that i learned how to get a general term for 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, etc along the way. the general term for this is:

$\displaystyle d_{n}=\left\lfloor\sqrt{2n}+\frac{1}{2}\right\rflo or$ - Jul 5th 2009, 08:38 AMkkoutsothodoros
- Jul 5th 2009, 08:42 AMmalaygoel
- Jul 5th 2009, 08:56 AMkkoutsothodoros
Consider 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,etc

where the integer m appears m times in the sequence. using the fact that

$\displaystyle 1+2+3+...+n=\frac{n(n+1)}{2}\$ we have $\displaystyle \frac{m(m-1)}{2}\$ < n <= $\displaystyle \frac{m(m+1)}{2}\$

this inequality leads to:

m < $\displaystyle \sqrt{2n}+\frac{1}{2}\$ <= m + 1 which is equivalent to what we are looking for. - Jul 5th 2009, 08:59 AMmalaygoelQuote:

Originally Posted by**kkoutsothodoros**

Now, how do you get this

Quote:

m < $\displaystyle \sqrt{2n}+\frac{1}{2}\$ <= m + 1

- Jul 5th 2009, 09:36 AMkkoutsothodoros
sorry again, this part is a little tricky and i found an error.

multiply $\displaystyle \frac{m(m-1)}{2}\$ < n <= $\displaystyle \frac{m(m+1)}{2}\$ by 2 to get $\displaystyle m^2 - m < 2n <=m^2 + m$. now add 1/4 to both sides and factor to get $\displaystyle m^2 - m + 1/4< 2n + 1/4 <=m^2 + m + 1/4$ or $\displaystyle (m-1/2)^2 < 2n + 1/4 <= (m + 1/2)^2$. take the square root of everything and we get $\displaystyle (m-1/2) < \sqrt{2n + 1/4}\ <= (m + 1/2)$.

add 1/2 to both sides and we get $\displaystyle m < \sqrt{2n + 1/4}\ + 1/2 <= m + 1$.

i just checked it in Excel and the 1/4 is not affecting the result, luckily. it still works if we start with n = 0.