I am trying to derive the duration of a perpetual bond with coupon $c$ in two ways:
$$D=-\frac{\frac{\partial P}{\partial r}}{P},$$ $$P=\frac{c}{r}$$ $$\Rightarrow D = -\frac{-\frac{c}{r^2}}{\frac{c}{r}}=\frac{1}{r}$$
In the second approach, I want to derive the duration using the Macauley Duration (average PV-weighted time to maturity):
$$D=\sum_{t=1}^T \frac{c_t}{(1+r)^tP}\cdot t$$ $$\Rightarrow D=\sum_{t=1}^\infty\frac{ c\cdot t}{(1+r)^t\frac{c}{r}}=\sum_{t=1}^\infty\frac{ r\cdot t}{(1+r)^t}=r\sum_{t=1}^\infty\left(\frac{1}{1+r}\right)^t\cdot t$$ However, I am unable to show the convergence of this sum to $1/y$.
I came as far as to rewriting the sum as: $$S_m=\sum_{k=1}^mkx^k=\sum_{k=0}^{m-1}(k+1)x^{k+1}=x+x\sum_{k=1}^{m-1}kx^k+x\sum_{k=1}^{m-1}x^k.$$ $$\Rightarrow (1-x)S_m=x\frac {1-x^m}{1-x}$$
For $y>0$ we have $x=\dfrac1{1+r}<1$ and so the sum converges to $$\Rightarrow S_m=\frac {x}{(1-x)^2}=\frac {\dfrac1{1+r}}{(1-\dfrac1{1+r})^2}$$
$$\Rightarrow D=\frac {\dfrac{r}{1+r}}{(1-\dfrac1{1+r})^2}$$ However, I was unable to show the desired result $D=\frac{1}{r}$.
Can someone show the correct solution?