Let's call S
=z^-1 + 2z^-2+ 3z^-3 +...+nz^-n
divide now both LHS and RHS by z^-1:
S
/z^-1=1 + 2z^-1+ 3z^-2 +...+nz^-n+1
we can note that (2z^-1+ 3z^-2 +...+nz^-n+1) - (z^-1+ z^-2 +...+z^-n+1) = z^-1 + 2z^-2+ 3z^-3 +...+(n-1)z^-n+1
if we call now A = z^-1+ z^-2 +...+z^-n+1 then, summing it to both LHS and RHS:
S
/z^-1 A=1 + z^-1+ 2z^-2 +...+(n-1)z^-n+1
but z^-1+ 2z^-2 +...+(n-1)z^-n+1 = S(n-1)
and S(n-1) = S
- nz^-n thus:
S
/z^-A = 1 - S
- nz^-n
S
[1/z^-1 +1] = 1 + A -nz^-n from which: S
= (1 + A - nz^-n)/(1/z^-1 +1)
We have to find A:
A(k) = z^-1+z^-2+...+z^-k
dividing both sides by z^-1
A(k)/z^-1 = 1 + x^-1+x^-2+...+x^-n
as before x^-1+x^-2+...+x^-n = A(k-1) = A(k) - x^-k
after some passage:
A(k) = (x^-1 - x^-k-1)/(1 - x^-1)
we have to find A(n-1), that is k=n-1:
A=A(n-1) = (x^-1 - x^-n)/(1 - x^-1)
You have now to substitute this result in S
then to have the sum to inf just calculate lim(n-->inf) S
even if I paid attention, please check that all my passages are corrrect.
In any case the solution found by elwo06 is faster and more elegant.