# Capital Recovery Factor for Hyperbolic Discounting

I'm looking for a closed-form capital recovery factor when hyperbolic discounting is used. The Wikipedia article on hyperbolic discounting has this formula for the "present value of a series of equal annual cash flows in arrears discounted hyperbolically": $$V = P \frac{\ln(1 + kd)}{k}\ ,$$ but does not include any references.

If I attempt to sum the series $$\sum_{t=1}^n 1/(1 + rt)$$ however, WolframAlpha gives me a more complicated formula involving the digamma function.

I would appreciate a reference for this CRF, if one exists. A derivation would be a bonus. Thanks!

• It looks like they've just approximated the sum with an integral Jan 20, 2020 at 10:37
• The sum will not have an exact form. en.m.wikipedia.org/wiki/Harmonic_number. This pages gives some nice asymptotic expansions in the case $r=1$. The logarithm approximation is pretty good though. Jan 20, 2020 at 10:46
• @B.Martin, thank you, that's helpful. Do you have a reference for the log approximation? Jan 22, 2020 at 17:10
• I don't have any specific references, but replacing a sum with an integral is a relatively common approximation. And in this case, you will end up with a really good approximation for large $n$. It will basically just be off by a small constant. Also to be completely transparent, I'm not that familiar with the economics of this situation, and was just giving the equations a look at. Jan 23, 2020 at 3:24
• @B.Martin, ah of course, that makes sense with the intergral. Thanks for the help! If you want to summarise this is in an answer, I'd accept it. Jan 23, 2020 at 9:19

I'll just expand upon what I've said in the comments as an answer. The sum $$\sum_{t=0}^n \frac{1}{1+rt}$$ will not have a nice closed form. However what wikipedia has done is approximate this sum as an integral. The integral will just be $$\int_{t=0}^n \frac{dt}{1+rt}=\frac1r\ln(1+rt)|_0^n=\frac1r \ln(1+nr)$$ Which is basically what you've got without multiplying by the principle value, and relabelling some variables. Now something to watch out for is that this will be off by a small constant. If you look at the diagram below, there is some error area that the integral fails to pick up, so it will actually be slightly smaller. In this case however you know the error area is below 1 (because if you imagine sliding each error area into the first square (purely horizontally it will not fill it all up)