-2

Up-front: Met the snobs on Math overflow and thought I'd try here instead. I have three STEM degrees including math, and this is for work not school; just trying to save hours googling.

This equation diverges so I can't just solve it. What I'd like to find is a way to sum a finite # of terms to a specific number of decimal places (say 4). X is a variable, c is a decimal less than 1. Ultimately, I hope to create a formula that will just calculate what I want. Any ideas?

$$\sum_{n=1}^{\infty} xc(1+c)^{n-1}$$

Lance Roberts
  • 181
  • 2
    This is simply a geometric series. Using the elementary formula for geometric series, this is equal to $-x$ for $|1+c|<1$. The wikipedia article contains more information on how this would evaluate for a finite number of summands. – Andreas Tsevas Mar 06 '24 at 00:40
  • 1
    When it converges (when $|1+c|<1$) it converges to $-x.$ – Thomas Andrews Mar 06 '24 at 00:41
  • @AndreasTsevas $-x,$ not $x.$ – Thomas Andrews Mar 06 '24 at 00:42
  • @AndreasTsevas |1+c| is greater than 1. – Lance Roberts Mar 06 '24 at 00:42
  • 1
    The term $xc$ has no dependence on $n$, so take it out: $xc\sum_{n \geq 1} (1 + c)^{n-1}$. By "sum a finite number of terms" are you asking how to compute $\sum_{n=1}^N (1+c)^{n-1} = 1 + (1+c) + (1+c)^2 + \cdots + (1+c)^{N-1}$? – KCd Mar 06 '24 at 00:42
  • If $|1+c|>1,$ the series does not converge. @LanceRoberts – Thomas Andrews Mar 06 '24 at 00:42
  • @KCd yes, but to end up with say 4 decimal places that are solid – Lance Roberts Mar 06 '24 at 00:43
  • @ThomasAndrews, yes, exactly. I said in the question that I know it diverges. I just want to sum up some amount of terms to get to a reasonable number I can use instead of hand-iterating every time. – Lance Roberts Mar 06 '24 at 00:44
  • 3
    If you want a finite number of terms, ask for a finite number of terms. There is no real approximation for a divergent series. @LanceRoberts – Thomas Andrews Mar 06 '24 at 00:45
  • @KCd Thanks for pointing out about pulling xc out, not sure how I missed that. Since it looks like I won't get an answer, I can use that to simplify my hand calculating process and get something reasonable. – Lance Roberts Mar 06 '24 at 00:52
  • 3
    If you [edit] the question to clarify what you mean by your comment "some number of terms to get to a reasonable number" we might be able to help. Include several numerical examples that satisfy you, and explain why they do. While you are editing, delete the tacky first paragraph of the question. – Ethan Bolker Mar 06 '24 at 00:53

1 Answers1

1

When $r \not= 1$, $$ \sum_{n=1}^{N} r^{n-1} = 1 + r + r^2 + \cdots + r^{N-1} = \frac{r^N-1}{r-1} = \frac{1-r^N}{1-r}. $$ That is a standard formula for a finite geometric series.

KCd
  • 46,062