Why is $\sum\limits_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t} = \Delta D$? In a book on economics, I have read that $\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t} = \frac{r\Delta D}{r} = \Delta D$.
$t$ is the index of the current time period; $r$ represents the interest rate; $\Delta D$ is the change of debts.
Why can this equation be solved like this? Let’s say $r = 2$. Then, $\sum_{t = 1}^{\infty} \frac{2 \Delta D}{(1 + 2)^t}$ should result in something like $\frac{2\Delta D}{3}$, shouldn’t it?
What am I getting wrong here? Why does $\sum_{t = 1}^{\infty} \frac{r}{(1 + r)^t} = \frac{r}{r}$?
 A: $$\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t}=\\r\Delta D \sum_{t = 1}^{\infty} \frac{1}{(1 + r)^t}=\\ $$ now note that $$r>0 \to r+1>1 \\\to
0<\frac{1}{1+r}<1$$
now see whats the value of $\sum_{t = 1}^{\infty} \frac{1}{(1 + r)^t}$ let it's name be $S=\sum_{t = 1}^{\infty} \frac{1}{(1 + r)^t}$
$$S= \frac{1}{(1 + r)^1}+\frac{1}{(1 + r)^2}+\frac{1}{(1 + r)^3}+...$$multiply  $S$ by $\frac{1}{(1 + r)^1}$ so 
$$\frac{1}{(1 + r)^1}S=\frac{1}{(1 + r)^2}+\frac{1}{(1 + r)^3}+\frac{1}{(1 + r)^4}+...$$ no look at $S-\frac{1}{(1 + r)^1}S$ 
$$S-\frac{1}{(1 + r)^1}S=\\
(\frac{1}{(1 + r)^1}+\frac{1}{(1 + r)^2}+\frac{1}{(1 + r)^3}+...)-(\frac{1}{(1 + r)^2}+\frac{1}{(1 + r)^3}+\frac{1}{(1 + r)^3}+...) \\ \to S-\frac{1}{(1 + r)^1}S=\frac{1}{(1 + r)^1}$$simpllify
$$S(1-\frac{1}{(1 + r)^1})=\frac{1}{(1 + r)^1}\\
S(\frac{1+r-1}{(1 + r)^1})=\frac{1}{(1 + r)^1}\\\to \\
S=\frac{\frac{1}{(1 + r)^1}}{\frac{r}{(1 + r)^1}}=\frac1r$$now look at begining
$$\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t}=\\r\Delta D \sum_{t = 1}^{\infty} \frac{1}{(1 + r)^t}=\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t}=\\r\Delta D S=\\r\Delta D \times\color{red} {\frac1r}\\=\Delta D $$
A: The main theme here is the geometric series expansion
\begin{align*}
\sum_{t=0}^\infty q^t=\frac{1}{1-q}\qquad\qquad |q|<1
\end{align*}
Since OPs series starts with index $t=1$ we consider
\begin{align*}
\sum_{t=1}^\infty q^t&=\left(\sum_{t=0}^\infty q^t\right)-1\\
&=\frac{1}{1-q}-1\\
&=\frac{q}{1-q}\qquad\qquad |q|<1\tag{1}
\end{align*}

Using (1) we can transform OPs series
  \begin{align*}
\sum_{t=1}^\infty\frac{1}{(1+r)^t}&=\sum_{t=1}^\infty\left(\frac{1}{1+r}\right)^t
=\frac{\frac{1}{1+r}}{1-\frac{1}{1+r}}\\
&=\frac{1}{r}\tag{2}
\end{align*}
  valid for $\left|\frac{1}{1+r}\right|<1$.

Putting all together we conclude from (2) the following is valid for 
$\left|\frac{1}{1+r}\right|<1$:

\begin{align*}
\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t}&=r\Delta D\cdot\sum_{t=1}^\infty \frac{1}{(1+r)^t}\\
&=r\Delta D\cdot\frac{1}{r}\\
&=\Delta D\tag{3}
\end{align*}

With respect to the example $r=2$: Since $\left|\frac{1}{1+2}\right|=\frac{1}{3}<1$ we obtain according to (2)
\begin{align*}
\sum_{t = 1}^{\infty} \frac{2\Delta D}{(1 + 2)^t}&=2\Delta D\cdot\sum_{t = 1}^{\infty}\left(\frac{1}{3}\right)^t\\
&=2\Delta D\cdot\frac{1}{2}\\
&=\Delta D
\end{align*}
in accordance with (3).
A: $$r > 0 \implies \frac{1}{1+r} < 1$$
then use GP.
For $|x| < 1$,
$$\sum_{i=0}^{\infty}x^i = \frac{1}{1-x}$$
