How to simplify $\sum_{i=t+1}^{\infty}\frac{1}{\left(1+r\right)^{i}}$? Starting from the following series:
$$\sum_{i=1}^t \dfrac{C}{\left( 1+r \right)^i}$$
one has that for $\mid\dfrac{1}{1+r}\mid<1$:
$$\sum_{i=1}^t \dfrac{C}{\left( 1+r \right)^i}=\underbrace{C\sum_{i=1}^{\infty}\dfrac{1}{\left(1+r\right)^{i}}}_{\text{converges to }\dfrac{C}{r}}-C\sum_{i=t+1}^{\infty}\dfrac{1}{\left(1+r\right)^{i}}=\dfrac{C}{r}-C\sum_{i=t+1}^{\infty}\dfrac{1}{\left(1+r\right)^{i}}$$
$$=C\left(\dfrac{1}{r}-\color{red}{\sum_{i=t+1}^{\infty}\dfrac{1}{\left(1+r\right)^{i}}}\right)$$

How could I simplify the series in $\color{red}{\text{red}}$?
 A: Let $q\in\{ x\in\mathbb R \mid \lvert x \rvert <1, x\neq 1 \}$ and $n,m\in\mathbb N$. Then $$\sum_{k=m}^n cq^k=\frac{c(q^m-q^{n+1})}{1-q}.$$
For $q=1$ the series is equal to $c(n-m+1)$. The proof is analogous to the well-known case $m=1, n=\infty$.
Your red series satisfies $m=t+1,n=\infty, c=1$ and $q=\frac{1}{r+1}$. I assume that $\lvert\frac{1}{r+1}\rvert<1$. Then plugging in these values into the above formula yields $$\frac{(\frac{1}{r+1})^{t+1}}{1-\frac{1}{r+1}}.$$
A: It seems that you are very familiar with geometric series in the infinite case, but maybe less so in the finite case.  Let's start with this finite sum:
$$
\sum_{i=1}^tx^i=x+x^2 +x^3 +...+x^t 
$$
The trick here is to factor out $x$ from all but the first term like this:
$$
\sum_{i=1}^tx^i=x+x(x^1 +x^2 +...+x^{t-1}) 
$$
Notice how the expression in parentheses is almost the original series! It is only missing the $x^t$ term. So we can write:
$$
\sum_{i=1}^tx^i = x + x(\sum_{i=1}^tx^i - x^t)
$$
Now its just a matter of algebra to show that:
$$
\sum_{i=1}^tx^i = \frac{x-x^{t+1}}{1-x}
$$
If we apply this to your equation it is clear that
$$
\sum_{i=t+1}^\infty \frac{1}{(1+r)^i} = \frac{1}{r}-\frac{\frac{1}{1+r}-\left(\frac{1}{1+r}\right)^{t+1}}{1-\frac{1}{1+r}}
$$
Simplifying
$$
\sum_{i=t+1}^\infty\frac{1}{(1+r)^i} = \frac{1}{r}\left(\frac{1}{1+r}\right)^t
$$
A: A variation.

We obtain
\begin{align*}
\color{blue}{\sum_{i=t+1}^{\infty}\frac{1}{(1+r)^i}}
&=\sum_{i=0}^\infty\frac{1}{(1+r)^{i+t+1}}\tag{1}\\
&=\frac{1}{(1+r)^{t+1}}\sum_{i=0}^\infty\left(\frac{1}{r+1}\right)^i\tag{2}\\
&=\frac{1}{(1+r)^{t+1}}\cdot\frac{1}{1-\frac{1}{1+r}}\tag{3}\\
&\,\,\color{blue}{=\frac{1}{r(1+r)^{t}}}
\end{align*}

Comment:

*

*In (1) we shift the index to start with $i=0$. In order to compensate this shift we have to replace each occurrence of $i$ with $i+t+1$ within the scope of the sum.


*In (2) we factor out terms which do not depend on $i$.


*In (3) we apply the finite geometric sum formula and simplify afterwards.
