Question about a summation problem from finance I'm studying a journal article about finance and I have trouble understanding how the author reach a result.
The equation he begins with is:
$$VTS_0=TD_0+T\Sigma_1^\infty PV_0[ΔD_t] \tag 1$$
Then he is stating that:
$$PV_0[\Delta D_t]=\Delta D_0\frac{(1+g)^t}{(1+R_f)^t} \tag 2$$
And by substituting he reach the result:
$$VTS_0=TD_0+T\frac{gD_0}{R_f-g} \tag 3$$
My problem is that I can't understand which summation properties he uses to reach the result from $(1)$ to $(3)$ using $(2)$.
 A: $$
VTS_0 = TD_0 + T\sum_{t=1}^{\infty}PV\left[\Delta D_t\right]
$$
where 
$$
T\sum_{t=1}^{\infty}PV\left[\Delta D_t\right] = T\sum_{t=1}^{\infty}\Delta D_0\left(\dfrac{1+g}{1+R_f}\right)^t 
$$
Concentrating on the sum
$$
\sum_{t=1}^{\infty}\left(\dfrac{1+g}{1+R_f}\right)^t = \left(\dfrac{1+g}{1+r_f}\right)\sum_{k=0}^{\infty}\left(\dfrac{1+g}{1+R_f}\right)^k 
$$
this sum is equal to
$$
\dfrac{1+g}{R_f-g}
$$
putting it all together I find
$$
VTS_0 = TD_0 + T\Delta D_0 \left(\dfrac{1+g}{R_f-g}\right)
$$
so if
$$
\Delta (1+g) = g
$$
then maybe thats how. But could you clarify what $\Delta$ is
A: I think he uses the following property:
$$\sum_{t=1}^{\infty}PV_0[\Delta D_t] = \sum_{t=1}^{\infty} \Delta D_0
\frac{(1+g)^t}{(1+R_f)^t} = \frac{g D_0}{R_f - g}.$$
At this point it seems clear that the author does not use summation properties per se. Instead, he derives the last part as a sum of an infinite series $\sum_{t=1}^{\infty} \Delta D_0
\frac{(1+g)^t}{(1+R_f)^t}$. It should converge if $R_f > g$. 
However, at his point I don't think I can give a more precise answer as to how exactly calculate the resulting sum of the given infinite series.
