Show that $\frac{1}{1+x}H(\frac{x}{1+x})=\sum^\infty_{k=0}[\Delta^kh_0]x^k$ For a sequence $\{h_n\}_{\geq 0}$, let $H(x)=\sum_{n\geq0}h_nx^n$.  Show that:
$$\frac{1}{1+x}H(\frac{x}{1+x})=\sum^\infty_{k=0}[\Delta^kh_0]x^k$$
What I did was that by proving the $$\Delta^k h_o=\sum^k_{j=0}(-1)^{k-j}{k \choose j}h_j$$
But no clue how to continue.
Help appreciated.

What we seek to show here is that
$$\sum_{k=0}^n {n\choose k} (-1)^{n-k} h_k
= [z^n] \frac{1}{1+z} H\left(\frac{z}{1+z}\right)$$
where $$H(z) = \sum_{q\ge 0} h_q z^q.$$
The RHS is given by the integral
$$\frac{1}{2\pi i}\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{1+z} H\left(\frac{z}{1+z}\right) \; dz
\\ = \frac{1}{2\pi i}\int_{|z|=\epsilon} \frac{1}{z^{n+2}}
\frac{z}{1+z} H\left(\frac{z}{1+z}\right) \; dz.$$
Now put $$\frac{z}{1+z} = u \quad\text{so that}\quad
z = \frac{u}{1-u}$$
and $$dz = \frac{1}{1-u} - \frac{u}{(1-u)^2} (-1) \; du
= \frac{1}{(1-u)^2} \; du.$$
We get for the integral
$$\frac{1}{2\pi i}\int_{|u|=\epsilon} \frac{(1-u)^{n+2}}{u^{n+2}}
\times u \times H\left(u\right) \; \frac{1}{(1-u)^2} \; du
\\ = \frac{1}{2\pi i}\int_{|u|=\epsilon} \frac{(1-u)^{n}}{u^{n+1}}
\times H\left(u\right) \; du.$$
This is $$[u^n] H(u) (1-u)^n$$
which evaluates to
$$\sum_{k=0}^n h_k {n\choose n-k} (-1)^{n-k}
= \sum_{k=0}^n {n\choose k} (-1)^{n-k} h_k$$
by inspection.
