Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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.
Well, I can rearrange the sum but I don't think that I've made it better. Anyway, here's what I've got.
$\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} $
And that's it.
