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.

When I fed it to Alpha the numerator comes out $(1+r+rT_2)(1+r)^{-1-T_2}-(1+rT_1)(1+r)^{-T_1}$.  If you want this to be less than zero, the inequality comes out in the opposite sense from yours.  If you divide by $(1+r)^{T_1}$ you get $(1+r+rT_2)\lt(1+rT_1)(1+r)^{1+T_2-T_1}$, then if you use $(1+r)^{1+T_2-T_1}\gt 1+r(1+T_2-T_1)$ you can get the RHS is greater than $(1+rT_1)(1+r(1+T_2-T_1))$ which is still greater than the LHS.  As all the steps are reversible, we know the original is less than $0$
Added:  We want to prove $(1+r+rT_2)\lt(1+rT_1)(1+r)^{1+T_2-T_1}$.  As $(1+r)^x \gt 1+xr$, $(1+rT_1)(1+r)^{1+T_2-T_1} \gt (1+rT_1)(1+r)(1+T_2-T_1)=(1+r+rT_1+r^2T_1)(1+T_2-T_1)$
$=1+r+rT_1+r^2T_1+T_2+rT_2+rT_2T_1+r^2T_2T_1-T_1-rT_1-rT_1^2-r^2T_1^2$
$=1+r+rT_2+(r^2T_1+(T_2-T_1)(1+r+r^2))\gt1+r+rT_2$
