Summation of infinite series, where difference in consecutive denominator forms an A.P.

Question

What is the sum of an infinite series where

1. each term can be written as $\frac{p}{q}$, where p=1 always

2. the difference between 2 consecutive denominators forms an A.P.

For example $\dfrac{1}{2}$, $\dfrac{1}{6}$, $\dfrac{1}{12}$, $\dfrac{1}{20}$, $\dfrac{1}{30}$ ......

Here,

$6-2=4$
$12-6=6$
$20-12=8$
$30-20=10$
.
.
.
.
.

where, 4,6,8,10.... form an arithmetic progression

Note: I would prefer if this question be answered using only elementary math, however, other solutions would also be appreciated. I have no knowledge of using limits

Answer 1

Hint. Your denominators $$ 2,6,12,20,30,\ldots $$ are such that $$ d_{n+1}-d_n=2(n+1),\quad d_1=2, \quad n=1,2,3,\cdots, $$ giving, by telescoping, $$ d_n=n(n+1), \quad n=1,2,3,\cdots. $$ Thus your series (a telescoping one) rewrites

$$ \lim_{N \to \infty}\sum_{n=1}^N\frac1{n(n+1)}=\lim_{N \to \infty}\sum_{n=1}^N\left(\frac1n-\frac1{n+1}\right)=\lim_{N \to \infty}\left(1-\frac1{N+1}\right)=\color{red}{1}. $$