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Please correct improper notation/terminology

$$\sum_{k=0}^{n-1} ar^k$$. $$\sum_{k=1}^{n} ar^{k-1}$$

As far as I can tell these both represent the same thing. It's the partial sum {$S_n$} where the starting index is {$term_1$} with the initial value {$a$}, and the ending index is {$term_n$}.

First question Is the difference between a partial sum vs using the summation notation {$\sum$} the fact that a partial sum always starts at {$term_1$} and adds sequentially left-to-right until {$term_n$}, as opposed to being able to start at any {$term_x$} using {$\sum$} while adding in any specified order?

Second question In {$\sum_{k=0}^{n-1} ar^k$}, {$k=0$} and {$n-1$} are implied to be direct exponent values of {$r$} for {$term_1$} and {$term_n$} shown respectively as {$ar^k$} and {$ar^{n-1}$}.

In contrast, {$k$} and {$n$} in {$\sum_{k=1}^{n}ar^{k-1}$} are implied to be direct term values - which can be shown as {$term_k$} and {$term_n$}. Variables {$ar^k$} and {$ar^{k-1}$} seem to be the modifiers responsible for this change.

  • What are the rules in notational convention governing this behaviour?
  • Are {$ar^k$} and {$ar^{k-1}$} variables representing initial value {$a$} (i.e. the value of {$term_1$}); are they notations representing the geometric sequence; or are they a convention used to dictate how {$k$} and {$n$} should be interpreted/used i.e. is {$k-1$} telling you to "shift" the entire progression to the side by one term?
    • Slightly different version of the last point: what exactly does {$k-1$} mean, particularly in regards to {$n$}? There's obviously some rule here because the notation changed from {$n-1$} to {$n$}.
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  • $\begingroup$ The $\Sigma$ is a Greek Sigma and is called the summation operator. I would call the various expressions equal to other expressions equalities leading to formulae for geometric series $\endgroup$ – Henry Apr 4 '16 at 23:11
  • $\begingroup$ What do you mean by "Why isn't a static value of term - 1,2,3...."? Where do you suggest these terms be used? $\endgroup$ – NoChance Apr 5 '16 at 12:57
  • $\begingroup$ @nochance k and n below and above sigma, these are supposed to represent the first and last terms (ie beginning and end of the sequence of partial sums) is it not ? $\endgroup$ – user5948022 Apr 5 '16 at 12:59
  • $\begingroup$ Your Question is difficult to understand. It seems you don't see any value to using variables in the summation limits rather than constants. This is really fundamental, to be able to make general statements (and prove them) rather than particular ones. Facility with changes of variables begins with high school algebra and continues to be an important skill throughout our mathematical journeys. $\endgroup$ – hardmath Apr 5 '16 at 13:20
  • $\begingroup$ The images have a lot of text and formulas that seem to be irrelevant to your question. As near as I can make out, you want to know why there isn't a universal convention that the summation index always starts at $1$ (or a convention that it always starts at $0$), instead of one author choosing $0$ and another choosing $1$. My guess is you don't object to $n$ as the last index. (The word "static", however, makes that guess very uncertain. I don't know what you think that word means in this context.) $\endgroup$ – David K Apr 5 '16 at 13:28
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I'll start from the bare basics.

Given a sequence $a_n$ (for example, the sequence of even natural numbers $a_1=2,a_2=4,\dots$).

We sometimes define a "sequence of partial sums" as $s_n=\sum_{k=1}^n a_k$.

Is it a convention that we should start at $1$? Well, no.

I like to start them at $0$ for example, perhaps some other author likes them starting at $31$, but that doesn't really change much. Anyone who writes something like "Let $s_n$ denote the sequence of partial sums of $a_n$" should make explicit what the starting index is (as a LOT of people either $0$ or $1$ as standard, but this is really arbitrary).

Remark: Even if we define $s_n$ as above, we can use this to denote a sum starting at any number we want.

How? Well, what does the expression $s_n-s_{x-1}$ mean (assume $x-1<n$)? From above, we see that

$$ s_n-s_{x-1}=\sum_{k=1}^n a_k-\sum_{k=1}^{x-1}a_k=(a_1+\cdots+a_x+\cdots+a_n)-(a_1+\cdots+a_{x-1})\\ =\sum_{k=x}^n a_k $$

Which is what you wanted!


Second question

I believe it's easier to see what you're doing this way: consider the sum

$$ \sum_{k=0}^n f(k) $$

This just says: Starting from $0$, up to the number $n$, evaluate $f$ at those points (i.e compute $f(0),f(1),...,f(n)$) and sum them up.

In your example, we have that $f(k)=ar^{k-1}$ (where $a$ denotes some constant).

From above, we see that your sum goes from $k=1$ to $k=n$, so all we have to do to compute it is calculate $f(1)=ar^{1-1}=a, f(2)=ar^{2-1},\dots$ and add them up.


Example of index shifting

Consider the following sum:

$$s=1+2+3+4+5+6+7+8+9+10$$

I believe you'll agree that I can write this as (see above, with $f(k)=k$) $$s=\sum_{k=1}^{10} k$$

Now I can write the same thing as $$ s=(2-1)+(3-1)+(4-1)+\cdots+(11-1) $$

Or, $$s=\sum_{k=2}^{11} (k-1)$$

Right? Well, from above you can see that these sums are really equal, but the indices and summands look pretty different. Is there any way to go from one to another easily?

Luckily for us, yes. What happens if we call $k-1$ "$u$"? Well, first we note that as $k$ moves from $2$ to $11$, the value of $u$ goes from $1$ to $10$, so we can write

$$ s=\sum_{u=1}^{10}u $$

Wow! Isn't this the first expression, with just a simple renaming of the variable? :)


Rearrangement of the terms

I'll give you an example of "switching the order of the sum".

Say we want to write $f(1)+f(2)+f(3)+\cdots+f(n)$ in sum notation? Well, simple enough, we've seen before that this is just $\sum_{k=1}^n f(k)$. But what if I want to add this terms in reverse order?

Well, I'll let you figure out if the following works $$ \sum_{k=1}^n f((n+1)-k) $$

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  • $\begingroup$ Great answer, the practical examples given progressively was just what I needed. Confirmed some of my pre-existing notions and cleared up other points of confusion. Also gave me new stuff I need to mull on so I might be back for some clarification. On a separate note, I've been on a math binge the last couple of days relying heavily on random tutorials and wikipedia; this has left something to be desired. You seem to have a knack for pedagogy, could you recommend any good sites/resources for someone who benefits from this kind of presentation? $\endgroup$ – user5948022 Apr 5 '16 at 21:13
  • $\begingroup$ Thanks for the compliment. Feel free to ask for clarification on anything. What are some particular topics you're interested in? I mostly come here when I'm learning new stuff, as the resources are almost endless (and there's lots of nice people willing to help you learn). Besides that, there's always books, but you'll have to be a bit more specific on what you're searching for. $\endgroup$ – YoTengoUnLCD Apr 5 '16 at 22:32
  • $\begingroup$ sorry I was talking about math in paricular. Any topic pre-calculus and on. Also anything geared towards the fundmental maths used in finance - I got into summation because of interest rates and inflation. Thanks $\endgroup$ – user5948022 Apr 6 '16 at 3:53
  • $\begingroup$ My last comment wasn't clear enough about the financial math I'm interested in. Better to say I want to learn about the fundamental math used in financial formulas. I could memorize (p-p((1+r)^-n))/r as the formula to adjust a mortgage for inflation but I wouldn't understand how that formula was derived. I could by learning about summation. It's a shot in the dark asking for such a specific resource so anything on pre-calculus would be great too. $\endgroup$ – user5948022 Apr 6 '16 at 5:37
  • $\begingroup$ I see, to be honest, I know nothing about finance (and the kind of mathematics used in that field), but if calc/pre-calc would help you, then I think you should check out Khan Academy (khanacademy.org). It's a pretty good resource for learning techniques and building up intuition. After that's acquired, you could grab a nice book to formalize your understandings. $\endgroup$ – YoTengoUnLCD Apr 6 '16 at 15:59

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