Why is $\sum\limits_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t} = \Delta D$?

Question

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 $\sum_{t = 1}^{\infty} \frac{r}{(1 + r)^t} = \frac{r}{r}$?

Answer 1

$$r > 0 \implies \frac{1}{1+r} < 1$$

then use GP.

For $|x| < 1$,

$$\sum_{i=0}^{\infty}x^i = \frac{1}{1-x}$$

Answer 2

$$\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t}=\\r\Delta D \sum_{t = 1}^{\infty} \frac{1}{(1 + r)^t}=\\ $$ 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 $$\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t}=\\r\Delta D \sum_{t = 1}^{\infty} \frac{1}{(1 + r)^t}=\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t}=\\r\Delta D S=\\r\Delta D \times\color{red} {\frac1r}\\=\Delta D $$

Answer 3

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).