Determine the present value of a perpetual annuity I want to determine the present value of a perpetual annuity, which will incur an interest payment of € 1 at the end of each year; 
A calculative interest rate $r$ is assumed. 
We are at the time $t = 0$, the first payout is in $t = 1$. 
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Could you maybe give me a hint about how we could calculate it? 
Is it maybe  $1+r\cdot t$ ? 
 A: I'll assume that you are the owner of the annuity, and that you receive the payment of 1 € at the end of each year. The issuer of the annuity will have to pay this amount. We assume that the interest rate is fixed at $r$, and we use this value for all calculations; this is the meaning of the term calculative interest rate. (In real life, interest rates will go up and down, which introduce uncertainty into the calculations.)
The present value of the first payment is $1\over 1+r$, since that is what the issuer will have to set aside today in order to make the payment at time $t=1$. The present value of the second payment is $1\over (1+r)^2$, which the issuer sets aside today for the payment of 1 € at time $t=2$. Continuing in the same minner, the present value of the perpetual annuity is the sum of the infinite geometric series
$${1\over 1+r}+{1\over (1+r)^2}+\dots={1\over 1+r}{1\over 1-{1\over 1+r}}={1\over r}$$
If you were to receive $A$ € each year, all terms above are multiplied by $A$, and the present value becomes $A\over r$, in agreement with the formula in Wikipedia.
A: Let denote the annuity denote as $a$ and the interest rate as $r$. If the annuity incur an interest of 1€ per year  then $a\cdot r=1$.
You receive an amount of a€ for $n$  periods. The present value of that payments is 
$PV=\frac{a}{q^n}\cdot \sum_{i=0}^{n-1} q^i=\frac{a}{q^n}\cdot \frac{1-q^n}{1-q} \quad$ with $q=1+r$
$=\frac{a}{q^n}\cdot \frac{1}{1-q}-\frac{a}{q^n}\cdot \frac{q^n}{1-q} =\frac{a}{q^n}\cdot \frac{1}{1-q}- \frac{a}{1-q}= \frac{a}{q^n}\cdot \frac{1}{1-q}+ \frac{a}{q-1}$
Now  $n$ goes to infinity.
$$\lim_{n\to \infty } \frac{a}{q^n}\cdot \frac{1}{1-q}+\lim_{n\to \infty }\frac{a}{q-1} $$
$=0+\frac{a}{q-1}$
with $q=1+r$ we get
$PV=\frac{a}{1+r-1}=\frac{a}r$
With $a\cdot r=1$
$PV=a^2$ or $PV=\frac1{r^2}$
