formula for a sequence

**kkoutsothodoros** · July 4th 2009, 10:24 AM

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..

**malaygoel** · July 5th 2009, 02:50 AM

I could get upto this:

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

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

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

someone, please silmpify

**kkoutsothodoros** · July 5th 2009, 06:46 AM

$d_{n}=\left\lfloor\sqrt{2n}+\frac{1}{2}\right\rflo or$

**kkoutsothodoros** · July 5th 2009, 07:12 AM

we get the general term of the sequence to be:

$a_{n}$ = $\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!!

**kkoutsothodoros** · July 5th 2009, 07:21 AM

i will probably get thrown off the site now!! well it was fun learning a little bit of latex!

**kkoutsothodoros** · July 5th 2009, 07:44 AM

oh yes, one more thing,

$a_{0}$ =0.

**malaygoel** · July 5th 2009, 07:55 AM

Originally Posted by kkoutsothodoros

we get the general term of the sequence to be:

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

I am not able to follow your answer...could you please give example for some values of n.

**kkoutsothodoros** · July 5th 2009, 08:22 AM

it does seem a little unclear. sorry, i am just learning how to use latex.

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

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

you should get

$s_{1}$ = 0
$s_{2}$ = .5
$s_{3}$ = 1
$s_{4}$ = 2/3
etc.

**kkoutsothodoros** · July 5th 2009, 08:24 AM

so $s_{n}$ 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

**kkoutsothodoros** · July 5th 2009, 08:35 AM

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:

$d_{n}=\left\lfloor\sqrt{2n}+\frac{1}{2}\right\rflo or$

**kkoutsothodoros** · July 5th 2009, 08:38 AM

Originally Posted by malaygoel

I am not able to follow your answer...could you please give example for some values of n.

is it more clear now?

**malaygoel** · July 5th 2009, 08:42 AM

Originally Posted by kkoutsothodoros

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:

$d_{n}=\left\lfloor\sqrt{2n}+\frac{1}{2}\right\rflo or$

your expression and my expession for d(given in post #2) are quite different. And they both work. Can you explain how you got your expression. mine had a simple logic.

**kkoutsothodoros** · July 5th 2009, 08:56 AM

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

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

this inequality leads to:

m < $\sqrt{2n}+\frac{1}{2}\$ <= m + 1 which is equivalent to what we are looking for.

**malaygoel** · July 5th 2009, 08:59 AM

Originally Posted by kkoutsothodoros

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

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

Upto here, everything is correct.

Now, how do you get this

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

**kkoutsothodoros** · July 5th 2009, 09:36 AM

sorry again, this part is a little tricky and i found an error.

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

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

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.

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