From a lecture:
It is required to prove that $ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $ using generating functions. It goes as follows:
The generating function $ g(x) = 1+2x+3x^2+4x^3...= \sum_{n=0}^{\infty} (n+1)x^n = \frac{1}{(1-x)^2}$
Now define $h(x)=g(x) \times \frac{1}{1-x}$ which using the geometric identity means $h(x)=g(x) \times \sum_{k=1}^\infty x^k$
Somehow they jump to $[x^{n-1}]h(x)=1+2+3+4+5+6... $ (this takes the coefficient of $x^{n-1}$).
How does this last step work? How can it be proved? Thanks.
I don't believe the rest of the proof is relevant. It is just this step which is unclear.
$h(x)$ has been written as the product of $g(x)$ and $1/(1-x)$. The n-th term of the series for which the generating function is a product of two generating functions is the convolution product of the terms in the two factors. Because the factors in the sequence from $1/(1-x)$ are all 1, the convolution (sum) looks simply like a sum of terms generated by $g(x)$.