Difference between 2 Sum of n terms of geometric series formulas

  • Thread starter Lebombo
  • Start date
144
0
Difference between 2 "Sum of n terms of geometric series" formulas

Notation A)

[tex]S_{n}= \sum_{k=0}^{n - 1} ar^{k} = ar^{0} + ar^{1} + ar^{2} +...+ ar^{n-1} = \frac{a(1-r^{n})}{1-r}[/tex]

Proof:

[itex] S_{n}= ar^{0} + ar^{1} + ar^{2} +...+ ar^{n-1}[/itex]
- [itex] r*S_{n}= ar^{1} + ar^{2} + ar^{3} +...+ ar^{n}[/itex]
___________________________________________________
[itex]S_{n} - rS_{n} = a - ar^{n}[/itex]

[itex]S_{n}(1-r)=a(1-r^{n}) \Rightarrow S_{n} = \frac{a(1-r^{n})}{1-r}[/itex]



Notation B)


[itex]S_{n+1} = \sum_{k=0}^{n} ar^{k} = ar^{0} + ar^{1} + ar^{2} +...+ ar^{n} = \frac{a(1-r^{n+1})}{1-r}[/itex]

Proof:

[itex]S_{n+1} = r^{0} + r^{1} + r^{2} +...+ r^{n}[/itex]
- [itex] r*S_{n+1} = r^{1} + r^{2} + r^{3}+...+ r^{n+1}[/itex]
___________________________________________
[itex]S_{n} - r*S_{n} = r^{0} - r^{n+1}[/itex]


[itex]S_{n}(1-r)=1-r^{n+1} \Rightarrow S_{n} = \frac{1-r^{n+1}}{1-r}[/itex]



So the 2 differences between the formulas I see from the start is:

1) Notation A includes the "a" terms while Notation B only includes the "r" terms. I assume this means that Notation B assumes a=1

2) General terms of [itex]S_{n}[/itex] in Notation A end with exponent [itex]ar^{n-1}[/itex] while General terms of [itex]S_{n+1}[/itex] in Notation B end with exponent [itex]r^{n}[/itex] . I assume the different notation is due to the emphasis put on how the series is described.

If emphasis of notation is placed on [itex]S_{n}[/itex], then you get [itex]ar^{0} + ar^{1} + ar^{2} +...+ ar^{n-1}[/itex] thus for the last term, you get the classic formula of "nth term of geometric sequence."

If emphasis of notation is placed on a "cleaner" upper bound on the sigma notation such as [itex]\sum_{n=1}^{n} ar^{k}[/itex] then you get the series in the form [itex]ar^{0} + ar^{1} + ar^{2} +...+ ar^{n}[/itex] so for the last term, you no longer have the classic formula of "nth term of geometric sequence."


_______

So in summary, what I can conclude when accounting for the differences between the 2 formulas are that the formula will change a little when a=1 and will change again when the exponent of the last term in the series is "n" instead of "n+1."


Are these notations, formulas, and ascertains about the notation differences correct? I'm just making wild guesses, so I'd be appreciative of any feedback that can correct the assumptions I've made, or to add to the accounting for the differences in the notations and formulas.
 
Last edited:
Well the difference is primarily in what you are doing each one =)
In your second proof the missing a constant isn't there because of as you said the proof assumes it would equal one. That's why even at the step below in the proof it hasn't been included though this series is fundamentally different from the step before it where a was included.
[itex]S_{n+1} = r^{0} + r^{1} + r^{2} +...+ r^{n}[/itex]
Also what you're doing in the second proof is summing the first n+1 terms and not the first n which is different. This confusion is in this step:
[itex]S_{n} - r*S_{n} = r^{0} - r^{n+1}[/itex]
It should look like:
[itex]S_{n+1} - r*S_{n+1} = r^{0} - r^{n+1}[/itex]​
So that you don't lose what youre doing you must keep the n+1 as the sub for S constant otherwise its misleading, Interestingly though I think the second proof is a proof of convergence but don't take my word for it ;)
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top