$1+1/(2+3)+1/(4+5+6)...$ What is the value of the sum of the series
$$\frac{1}{1}+\frac{1}{2+3}+\frac{1}{4+5+6}+\dotso\;?$$
And this:  
$$\frac{1}{1}+\frac{1}{2\cdot3}+\frac{1}{4\cdot5\cdot6}+\dotso\;?$$
 A: Note $a+(a+1)+\cdots+(b-1)+b= (b-a+1)(b+a)/2$. Set $b=n(n+1)/2,a=n(n-1)/2+1$:
$$(\circ)=\sum_{n=1}^\infty \left((n(n+1)/2-[n(n-1)/2+1]+1)\cdot\frac{n(n+1)/2+[n(n-1)/2+1]}{2}\right)^{-1}$$
$$=\sum_{n=1}^\infty\frac{2}{n(n^2+1)}=\sum_{n=1}^\infty\left(\frac{2}{n}-\frac{1}{n+i}-\frac{1}{n-i}\right)=\sum_{n=1}^\infty\int_0^12x^{n-1}-x^{n+i-1}-x^{n-i-1}dx$$
$$=\int_0^1\left(\sum_{n=1}^\infty x^{n-1}\right)(2-x^i-x^{-i})dx=\int_0^1\frac{1-x^i}{1-x}dx+\int_0^1\frac{1-x^{-i}}{1-x}dx$$
$$=\psi(1+i)+\psi(1-i)+2\gamma.$$
(Where $\psi$ is the digamma function.) That's the first series. Second one doesn't look so easy.
A: There is a simple formula for $\sum_{i=k}^n i = \frac{n(n+1)}{2}-\frac{(k-1)k}{2}$.
Now, the m-th denominator has $m$ terms, which means that there are $1+2+..+(m-1)=\frac{(m-1)m}{2}$ terms before the first of the denominator.
Hence, your n'th term is 
$$\frac{1}{\sum_{i=\frac{(n-1)n}{2}+1}^{\frac{(n-1)n}{2}+n} i}$$
Use the formula at the begining of the proof, and you'll probably end up with a standard telescopic sum....
Edit See Andre's comment below, this is not telescopic, so read the above comments as "how to reduce this series to a simpler "closed" form series".. The exact formula is calculated below by Andre.
