Assume $~n\in\mathbb{N}_{\geq3}~$
$$\underbrace{\color{red}{
\begin{align}
\sum_{x=n+1}^{+\infty}(0.6)\cdot(0.4)^{x-1}=(0.4)^n
\end{align}
}
}_{\text{I want to derive RHS from LHS}}
$$
I know the following formula which could be applied to derive the red eqn.
$$S_n = {a\left(1-r^n\right)\over(1-r)}$$
$S_n~$represents the sum of a geometric sequence where $~a~$ stands for an initial term and $~r~$ takes a common ratio, and $~n\in\mathbb{N}~$ holds.
The problem is that, the index which is on the shoulder of $~(0.4)~$ doesn't start with $~1~$ and my brain has stopped thinking.
Can anyone give me some hint(s)?
BTW, I've written this post from my smartphone so my reply may be late since it is hard to write MathJax codes using the phone so far.
ADD
$$\begin{align}
&\sum_{x=n+1}^{+\infty}(0.6)\cdot(0.4)^{x-1}\\&=
\end{align}$$
Best Answer
The formula for the sum of a (convergent) infinite geometric series is $$\frac{\text{FirstTerm}}{1-\text{CommonRatio}}$$ where FirstTerm is the first term of the series and CommonRatio is the ratio of each term (except the first) to the one before. The series is convergent if the absolute value of CommonRatio is less than one. In your example, $\text{FirstTerm}=(0.6)\cdot(0.4)^n$ and $\text{CommonRatio}=0.4$.
The formula for the sum of a finite geometric series with $\text{CommonRatio}\neq1$ is $$\frac{\text{FirstTerm}-\text{FirstMissingTerm}}{1-\text{CommonRatio}}$$ where FirstMissingTerm is the term that would come next if the series were to be continued.