Give a closed expression for the infinite sum:

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

(In reply to

answer by K Sengupta)

We can write the jth term of the iven sequence S(j) as:

1/ S(j) = Sum(m= 1 to j) (AM(T(j-1) + m)), where AM

denotes the Arithmetic Mean and T(r) = rth Triangular Number.

Thus, j/S(j)

= j*[j(j-1)/2] + j(j+1)/2

= j(j^2+1)/2, giving:

S(j) = 2/(j^2+1)

Thus, the reqired sum to be evaluated reduces to:

2* Sum(j= 1 To infinity)[1/(j^2 + 1]

= -2 + Sum(j= 0 To + infinity)[1/(j^2 +1)]

= -1 + Sum(j= - infinity To + infinity)[1/(j^2 +1)]

Now, it can be established by a combination of Fourier Expansion and Euler's Formula, that:

Sum(j = - infinity To + infinity)[1/(j^2 + b^2)]

= b*pi*coth(pi)

Consequently, substituting b = 1, we obtain the required som as:

pi*coth(pi) - 1

= 2.1533481(approx) , where :

coth(pi) = (e^(2pi) + 1)/(e^(2pi) - 1)