How to prove that the Derivative of the Annuity Present Value Factor is negative I work with the Annuity present value factor, which I want to differentiate with respect to r:
$$\sum_{t=T_1}^{T_2} (1+r)^{-t}$$ which equals $$\frac{(1 + r)^{1 - T_1} - (1 + r)^{-T_2}}{r}$$
If I use the derivation of the Sum-expression I can easily proof, that the derivative with respect to r ist <0. If i do the same with the other (but equal) term, after some transforming I am stuck with $$(1 + r)^{1 + T_2} \cdot (1 + r \cdot T_1) < (1 + r)^{T_1} \cdot (1 + r \cdot(1+ T_2))$$ at which point I don't see how to proceed. How to solve this inequation? There must be some math rules or relations I am missing.

EDIT: The conditions are $0<r<1,\quad T_1\geq 1 \quad T_1<T_2$

EDIT: I tried to follow @Ross's instructions. I then end up with $$\frac{(1+r\cdot(1+T_2))}{(1+r\cdot T_1)} < (1+r)^{1+T_2-T_1}$$
Unfortunately I still don't know how to proceed from there on.

Let $v = \frac{1}{1+r}$. Then the present value is $$v^{T_1}+v^2+ \cdots + v^{T_2}$$ $$=v^{T_1}(1+v^{2-T_1} + \cdots + v^{T_2-T_1})$$ $$= v^{T_1} \left(\frac{1-v^{T_2-T_{1}+1}}{1-v} \right)$$ $$= v^{T_1-1} \left(\frac{1-v^{T_2-T_{1}+1}}{v^{-1}-1} \right)$$
So the derivative with respect to $v$ is $$\frac{-(T_2+1)v^{T_2+1}+T_{2}v^{T_{2}+2}-(T_{1}-1)v^{T_{1}+1}+T_{1}v^{T_{1}}}{(v-1)^{2}v}$$
