The Sum of Inverse Triangle Numbers

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The Sum of Inverse Triangle Numbers

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 $n$th arrangement the "$n$th triangle number," and denote it $T_n$. Thus we have $T_1 = 1$, $T_2 = 3$, and $T_3 = 6$.

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

Second question (this one is a little harder): Does
$\sum_{n=1}^\infty\frac{1}{T_n}$
converge? If so, to what does it converge?

xander

xander
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Re: The Sum of Inverse Triangle Numbers

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?
Q.E.D. , or not?
Zapp
University

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Re: The Sum of Inverse Triangle Numbers

Zapp wrote: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. ;)

Zapp wrote: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

xander
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Joined: Fri Feb 11, 2011 12:14 am
Location: Sparks, NV, USA

Re: The Sum of Inverse Triangle Numbers

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
Q.E.D. , or not?
Zapp
University

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Joined: Thu Feb 10, 2011 3:49 pm

Re: The Sum of Inverse Triangle Numbers

The second part of this question is nice:
Spoiler! :
We have from part 1 that $T_n=\frac{1}{2}n(n+1)$. Reciprocating and rewriting by partial fractions, we get $\frac{1}{T_n}=2(\frac{1}{n}-\frac{1}{n-1})$. The sum will telescope, leaving just $2(\frac{1}{1}-\lim_{n \to \infty}\frac{1}{n+1})$, so the sum is just 2.

DeathRowKitty
Mathlete

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Re: The Sum of Inverse Triangle Numbers

DeathRowKitty wrote:The second part of this question is nice:
Spoiler! :
We have from part 1 that $T_n=\frac{1}{2}n(n+1)$. Reciprocating and rewriting by partial fractions, we get $\frac{1}{T_n}=2(\frac{1}{n}-\frac{1}{n-1})$. The sum will telescope, leaving just $2(\frac{1}{1}-\lim_{n \to \infty}\frac{1}{n+1})$, 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

xander
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Re: The Sum of Inverse Triangle Numbers

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

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Re: The Sum of Inverse Triangle Numbers

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

Prove it.

BGronin wrote: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...

xander
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