The Sum of Inverse Triangle Numbers

[xander]

Consider the following arrangements of dots:


In the first arrangement, there is one dot. In the second arrangement, there are three dots which form an equilateral triangle. This arrangement could be thought of as being the result of adding a row of two dots to the first arrangement. The third arrangement contains six dots, and can be thought of as the second arrangement, plus a row of three dots. We will call the number of dots in the th arrangement the "th triangle number," and denote it . Thus we have , , and .

First question (this one is relatively easy): Is there a general formula for ? If so, what is that formula?

Second question (this one is a little harder): Does

converge? If so, to what does it converge?

xander

[Zapp]

answer to the first question:

Spoiler! :

Tn equals the sum of the first n natural numbers, because Tn= Tn-1 + n. therefore Tn = n(n+1)/2

and I still don't know what converge means, so I've got no answer for the second^^
is converge related to a limit of some sort?

[xander]

answer to the first question:

Spoiler! :

Tn equals the sum of the first n natural numbers, because Tn= Tn-1 + n. therefore Tn = n(n+1)/2

Correct. Told you that was an easy one. ;)

and I still don't know what converge means, so I've got no answer for the second^^
is converge related to a limit of some sort?

http://mathworld.wolfram.com/ConvergentSeries.html

Basically, if the series approaches some limit, it converges.

xander

[Zapp]

ah, then we have:

Spoiler! :

let f(n)= 1/Tn= 2/(n² + n)
now if we compute: the sum from n=1 to infinity of 2/(n²+n) it will converge to 2

[DeathRowKitty]

The second part of this question is nice:

Spoiler! :

We have from part 1 that . Reciprocating and rewriting by partial fractions, we get . The sum will telescope, leaving just , so the sum is just 2.

[xander]

The second part of this question is nice:

Spoiler! :

We have from part 1 that . Reciprocating and rewriting by partial fractions, we get . The sum will telescope, leaving just , so the sum is just 2.

Yay! That's what I got, too!

Spoiler! :

But it took me way to long to work out the partial fractions. I need to practice that skill more often.

xander

[BGronin]

Is there a way to generalize this to any type of geometric figure numbers? i.e. reciprocal of the pentagonal numbers, hexagonal numbers, etc.?

[xander]

Spoiler! :

now if we compute: the sum from n=1 to infinity of 2/(n²+n) it will converge to 2

Prove it.

Is there a way to generalize this to any type of geometric figure numbers? i.e. reciprocal of the pentagonal numbers, hexagonal numbers, etc.?

Not that I am aware of. The solution that DeathRowKitty and I give takes advantage of the fact that [spoiler]the partial fractions expansion of the general term for triangle numbers telescopes.[/quote] The sums of other inverse figurate numbers will converge---they all grow faster than triangle numbers, so they are bounded above, and they are positive, so they are bounded below---but working out what they converge to is going to have to be handled on a case-by-case basis (unless you are aware of a general formula for figurate numbers?).

xander

EDIT: Hrm... Wolfram's Mathworld does give a general formula, but I am not sure that there is a good way to find the limit of that sum in the general case. I will have to give it some thought...