Cool series

The $n$ th denominator is of the form $\frac{n(n + 1)}{2}$ , so the series can be written as

$\displaystyle\sum_{k=1}^{\infty} \dfrac{2}{n(n + 1)} = 2*\sum_{k=1}^{\infty} (\dfrac{1}{n} - \dfrac{1}{n + 1}) = 2*1 = \boxed{2}$ ,

as this is a telescoping series.

We note that the series of triangle numbers is:

$\{1,3,6,10...\} = \{1, (1+2), (1+2+3), (1+2+3+4)...\}$

Its $n^{th}$ term is given: $T_n = \sum _{k=1} ^n {k} = \dfrac {n(n+1)}{2}$ Therefore, the sum of the series of reciprocals of triangle numbers is: $S = \sum _{n=1} ^\infty {\dfrac {2} {n(n+1)}} = 2 \sum _{n=1} ^\infty {\left( \dfrac {1} {n} - \dfrac {1}{n+1} \right)} = 2 \left( \dfrac {1}{1} \right) = \boxed {2}$

Firstly we use partial fractions: $\frac{1}{n(n+1)}$ = $\frac{a}{n}$ + $\frac{b}{(n+1)}$ $\Rightarrow$ a(n+1) + bn = 1. Let n=-1 gives b= -1 then a= 1 so $\displaystyle\sum_{r+1}^{\infty} \dfrac{2}{n(n+1)} = 2\times\displaystyle\sum_{r+1}^{\infty} (\dfrac{1}{n} - \dfrac{1}{n+1})$ $=2 ( (1-\frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) +...) = 2$

double x=0; int a=0; for(int i=1;i<=20000;i++) { x+= 1/(double)(i+a); a=i+a;}

  printf("%10.9lf",x);

(1 + 1/ 3) + (1/ 3 + 1/ 5)/ 2 + (1/ 5 + 1/ 7)/ 3 + (1/ 7 + 1/ 9)/ 4 + ...

1 + 1/ 3 x 3/ 2 + 1/ 5 x 5/ 6 + 1/ 7 x 7/ 12 + 1/ 9 x 9/ 20 + ...

1 + 1/ 2 + 1/ 6 + 1/ 12 + 1/ 20 + 1/ 30 + ...

1 + 1/ 2 + 1/ 2 x 1/ 3 + 1/ 3 x 1/ 4 + 1/ 4 x 1/ 5 + 1/ 5 x 1/ 6 + ...

1 + 1/ 2 + 1/ 2 - 1/ 3 + 1/ 3 - 1/ 4 + 1/ 4 - 1/ 5 + ...

2

If you take the numbers given, and add them together, you get 1 and 2/3. Adding the next number in the series gives you 1 and 5/7. If you keep doing this, the answer will get closer and closer to 2, and as the series is infinite, that's the answer.