<p>I'll just expand upon what I've said in the comments as an answer. The sum <span class="math-container">$$\sum_{t=0}^n \frac{1}{1+rt}$$</span> will not have a nice closed form. However what wikipedia has done is approximate this sum as an integral. The integral will just be <span class="math-container">$$\int_{t=0}^n \frac{dt}{1+rt}=\frac1r\ln(1+rt)|_0^n=\frac1r \ln(1+nr)$$</span> 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. 
 <a href="https://i.sstatic.net/UQt35.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UQt35.jpg" alt="enter image description here"></a>
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)</p>

<p><a href="https://i.sstatic.net/JsorF.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/JsorF.jpg" alt="enter image description here"></a> and above one half, (draw a line between the corners of each error area to make a triangle and each triangle is half the area of the corresponding rectangle, so the sum of all the triangles is half the area of the unit square)<a href="https://i.sstatic.net/E1llA.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/E1llA.jpg" alt="enter image description here"></a></p>

<p>So since the error is between 1 and a half, for large n it should be pretty insignificant, and this is a pretty good approximation. </p>

CC BY-SA 4.0
2020-01-23T22:56:10.260Z
2020-01-23T22:56:10.260Z
Ben Martin
<p>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":
<span class="math-container">$$
V = P \frac{\ln(1 + kd)}{k}\ ,
$$</span>
but does not include any references.</p>

<p>If I attempt to sum the series <span class="math-container">$\sum_{t=1}^n 1/(1 + rt)$</span> however, WolframAlpha gives me a more complicated formula involving the digamma function.</p>

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

CC BY-SA 4.0
2020-01-20T10:37:49.713Z
It looks like they've just approximated the sum with an integral
Ben Martin
CC BY-SA 4.0
2020-01-20T10:46:24.517Z
The sum will not have an exact form. https://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.
Ben Martin
CC BY-SA 4.0
2020-01-22T17:10:34.110Z
@B.Martin, thank you, that's helpful. Do you have a reference for the log approximation?
Anthony
CC BY-SA 4.0
2020-01-23T03:24:58.233Z
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.
Ben Martin
CC BY-SA 4.0
2020-01-23T09:19:21.380Z
@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.
Anthony
<reference-request><behavioral-economics>
Capital Recovery Factor for Hyperbolic Discounting
Anthony
# Capital Recovery Factor for Hyperbolic Discounting

(Asked by: username_0 on 2020-01-14)

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!

**Tags:** reference-request, behavioral-economics
> **Comments:**
> **username_1**: It looks like they've just approximated the sum with an integral  
> **username_1**: The sum will not have an exact form. https://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.  
> **username_0**: @B.Martin, thank you, that's helpful. Do you have a reference for the log approximation?  
> **username_1**: 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.  
> **username_0**: @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.  

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## 1 answers

### Answer by username_1 on 2020-01-23. Score: 1 | Accepted answer

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.
[![enter image description here](https://i.sstatic.net/UQt35.jpg)](https://i.sstatic.net/UQt35.jpg)
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)

[![enter image description here](https://i.sstatic.net/JsorF.jpg)](https://i.sstatic.net/JsorF.jpg) and above one half, (draw a line between the corners of each error area to make a triangle and each triangle is half the area of the corresponding rectangle, so the sum of all the triangles is half the area of the unit square)[![enter image description here](https://i.sstatic.net/E1llA.jpg)](https://i.sstatic.net/E1llA.jpg)

So since the error is between 1 and a half, for large n it should be pretty insignificant, and this is a pretty good approximation.


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