Give a closed expression for the infinite sum:

1/1 + 2/(2+3) + 3/(4+5+6) + 4/(7+8+9+10) + ...

Consider the *n*th term.

The last term in its denominator is the *n*th triangular number: ½n(n+1).

So the first term in its denominator is ½n(n-1) + 1.

The mean of these two terms is ½(n²+1).

As there are n terms in arithmetical progression in the denominator, the *n*th term equals n/[½n(n²+1)] = 2/(n²+1).

The sum from n = 1 to infinity of 2/(n²+1) may be evaluated using contour integration.

The answer is pi×coth(pi) - 1, where coth(x) = (e^{x} + e^{-x})/(e^{x} - e^{-x}).

(It strikes me this last step is quite difficult for a level 3, so I suspect Federico may have another solution up his sleeve!)