# Help with an equation for a blog post please

I don't know any calculus but I'm pretty sure it's what I need to figure out an equation I need for a blog post I'm working on.

Here's the blog post.
http://ratdog-justbecause.blogspot.com/2015/01/low-eroei-and-low-power-density-is.html
.
I think maybe what I'm looking for is the Σ as n approaches infinity for 1/xn. With n starting out as 0. I don't really know for sure though. Any help is appreciated. Thanks.

Last edited:

Mark44
Mentor
I don't know any calculus but I'm pretty sure it's what I need to figure out an equation I need for a blog post I'm working on.

Here's the blog post.
http://ratdog-justbecause.blogspot.com/2015/01/low-eroei-and-low-power-density-is.html
.
I think maybe what I'm looking for is the Σ as n approaches infinity for 1/xn. With n starting out as 0. I don't really know for sure though. Any help is appreciated. Thanks.
You have a section titled "Now for some math" but all I see are a bunch of numbers that you took out of a spreadsheet, and no indication of what these numbers mean or how you got them.

You have a section titled "Now for some math" but all I see are a bunch of numbers that you took out of a spreadsheet, and no indication of what these numbers mean or how you got them.
Sorry, I was a bit sloppy. It just seems so obvious to me.

EROEI describes the energy that is needed to be invested (i.e. used) in order to get more energy. If the EROEI is 2 then one unit (of some unit of energy) invested will get you 2 of that unit. If solar is to become a permanent thing then maintaining any given area of it will take energy. The amount of energy is determined by it's EROEI. Lets say that the EROEI is 2. Then maintaining 1m2 of it would take 1/2m2. That 1/2m2 require 1/4m2 and so on. The sum of all these works out to some finite number. The equation for that number was what I was looking for. For the spread sheet I carried out a number of the operation discussed above on the number 1. I used the number one as a place holder for all areas. I did this to create a number that I could multiply any given area by in order to find out the total area needed.

I recently talked to my brother about this. He told me what I'm looking for is called the Geometric series, and the equation is 1/1-(1/x). Sorry Should have asked him first sense I know he's good at math while I mostly go of instinct. I'm going to go fix my blog post. Thanks for your help.

Last edited:
Mark44
Mentor
Sorry, I was a bit sloppy. It just seems so obvious to me.

EROEI describes the energy that is needed to be invested (i.e. used) in order to get more energy. If the EROEI is 2 then one unit (of some unit of energy) invested will get you 2 of that unit. If solar is to become a permanent thing then maintaining any given area of it will take energy. The amount of energy is determined by it's EROEI. Lets say that the EROEI is 2. Then maintaining 1m2 of it would take 1/2m2. That 1/2m2 require 1/4m2 and so on. The sum of all these works out to some finite number. The equation for that number was what I was looking for. For the spread sheet I carried out a number of the operation discussed above on the number 1. I used the number one as a place holder for all areas. I did this to create a number that I could multiply any given area by in order to find out the total area needed.

I recently talked to my brother about this. He told me what I'm looking for is called the Geometric series, and the equation is 1/1-(1/x). Sorry Should have asked him first sense I know he's good at math while I mostly go of instinct. I'm going to go fix my blog post. Thanks for your help.
Yes, it's a geometric series.
$$\sum_{n = 0}^{\infty} \frac{1}{2^n} = 1 + \frac{1}{2^1} + \frac{1}{2^2} + ... + \frac{1}{2^n} + ... = 2$$

Yes, it's a geometric series.
$$\sum_{n = 0}^{\infty} \frac{1}{2^n} = 1 + \frac{1}{2^1} + \frac{1}{2^2} + ... + \frac{1}{2^n} + ... = 2$$
Thanks!