# 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.

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If you put your equations between dollar signs for inline math and double dollar signs for displayed math, you can use LaTeX, given that you are familiar with that. EDIT: If you try to edit your question, you can see how it works. –  Raeder Apr 27 '11 at 15:24
@Raeder: Thanks, that's really convenient :) –  meep.meep Apr 27 '11 at 15:29

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 $(1+r)^x \gt 1+xr$, $(1+rT_1)(1+r)^{1+T_2-T_1} \gt (1+rT_1)(1+r)(1+T_2-T_1)=(1+r+rT_1+r^2T_1)(1+T_2-T_1)$

$=1+r+rT_1+r^2T_1+T_2+rT_2+rT_2T_1+r^2T_2T_1-T_1-rT_1-rT_1^2-r^2T_1^2$

$=1+r+rT_2+(r^2T_1+(T_2-T_1)(1+r+r^2))\gt1+r+rT_2$

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Thanks for your help. Unfortunately I still don't get your last steps (I've edited my question). Could you be so kind to explain how you get to your last 2 expressions and why the "greater than" conditions apply? That would be great... –  meep.meep Apr 27 '11 at 18:15
Thanks again, now I understood it all :) –  meep.meep Apr 28 '11 at 9:02
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 $$\frac{-(T_2+1)v^{T_2+1}+T_{2}v^{T_{2}+2}-(T_{1}-1)v^{T_{1}+1}+T_{1}v^{T_{1}}}{(v-1)^{2}v}$$
Thanks... The thing is, I need the proof that this derivation is $<0$. –  meep.meep Apr 27 '11 at 18:17