# Sum of a Geometrico-geometric series

Is there a closed form solution for the sum of an infinite geometric series, where the growth from element n to element n+1 converges to zero for $$n\rightarrow\infty$$?

Specifically, I'm looking for a closed formula for $$\sum_{t=0}^\infty\left(1+a\phi^t\right)^t\omega^t$$ where $$\omega<\phi<1$$ as well as $$a<1$$.

Alternatively, a solution for $$\sum_{t=0}^\infty\left(1+\frac at\right)^t\omega^t$$ would also be great.

Thanks a lot for looking into this.

$$\begin{array}\\ S(a, b, c) &=\sum_{t=0}^\infty\left(1+ab^t\right)^tc^t\\ &=\sum_{t=0}^\infty\sum_{k=0}^t\binom{t}{k}a^kb^{tk}c^t\\ &=\sum_{k=0}^{\infty}\sum_{t=k}^{\infty}\binom{t}{k}a^kb^{tk}c^t\\ &=\sum_{k=0}^{\infty}a^k\sum_{t=k}^{\infty}\binom{t}{k}(b^{k}c)^t\\ &=\sum_{k=0}^{\infty}\dfrac{a^k}{k!}\sum_{t=k}^{\infty}\dfrac{t!}{(t-k)!}(b^{k}c)^t\\ &=\sum_{k=0}^{\infty}\dfrac{a^k}{k!}\sum_{t=0}^{\infty}\dfrac{(t+k)!}{t!}(b^{k}c)^{t+k}\\ &=\sum_{k=0}^{\infty}\dfrac{a^k(b^kc)^k}{k!}\sum_{t=0}^{\infty}\dfrac{(t+k)!}{t!}(b^{k}c)^{t}\\ &=\sum_{k=0}^{\infty}\dfrac{a^k(b^kc)^k}{k!}k!(1-b^kc)^{-k-1}\\ &=\sum_{k=0}^{\infty}\dfrac{a^k(b^kc)^k}{(1-b^kc)^{k+1}}\\ \end{array}$$