Summation Indices - How to interpret the zero index?

Question

A very quick background - I'm a programmer by trade that has decided to try and properly relearn the math I never fully grasped 10 years ago in school. I can "perform calculations", and am rather proficient at fortran, but at no point would I ever claim that I have any mathematical maturity. My current goal is remedy that. I'm currently working my through Apostol's Calculus, and finding it remarkable. However, there are a few problems I run in to where I'm struggling to understand the proper way to "look" at something - sometimes the proofs seem almost too easy and it feels as if I'm doing these wrong.

This is probably the simplest question ever asked on this board, but it's important to me. This question is in an unanswered Problem set, and I want to make sure I understand it correctly. Thanks in advance for taking the time.

Prove: $$\sum_{k=1}^n1 = n$$ Beginning: $$\sum_{k=1}^n1 = \sum_{k=1}^na_k$$ where each $a_k = 1$. We would like to utilize the Telescopic Property to simplify the sum. By the Telescopic Property: $\sum_{k=1}^n(a_k - a_{k-1}) = a_n - a_0$, let us rewrite $$\sum_{k=1}^n=a_k$$ to $$\sum_{k=1}^n(1-(1-1)) = \sum_{k=1}^n(1 - 0)$$ giving $$\sum_{k=1}^x(1-0) = n - 0 = n$$

Is this correct? My confusion comes from the term $a_0$. From my programming background, everything is zero indexed so I'm having a hard time reconciling $a_0$ with being zero in this case. I've spent several hours re-reading the previous pages in the textbook, and subscripts of zero aren't used . . . everything starts with an index of 1 (I'm assuming because we are dealing with the inductive set only at this time). Thanks!

Answer 1

Part of the proof seems to be confused (or something different is about to be shown). To make use of the telescoping property $$ \sum_{k=1}^n 1 = \sum_{k=1}^n (a_k-a_{k-1})=a_n-a_0,$$ you should not have $a_k=1$, but rather $a_k-a_{k-1}=1$. For example if we let $a_k=k+42$ for all $k$, then we indeed get $a_k-a_{k-1}=1$ for all $k$ and find quite unsurprisingly that $$ \sum_{k=1}^n 1=a_n-a_0=(n+42)-42=n.$$

Answer 2

I am not sure what level of rigor your problem set requires of you. As such, I can try to answer your question on two levels, one intuitive and the other very formalistic.

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We begin intuitively. Translating the summation

$$\sum_{k=1}^n1 = n$$

to English gives us 'the sum of $n$ ones, all considered simultaneously'. We would immediately recognize the answer as $n$.

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Let us now veer towards a more formalistic discussion that may seem crazily rigid at times.

What your source text calls the Telescoping Property is actually part of the very definition of addition and of the counting ('natural') numbers themselves. That is, we could translate the summation in question to English as 'the sum of $n$ ones, added sequentially'.

To see how that translation corresponds to your summation, we need to first forget whatever we think we know about numbers and addition. Starting from this empty slate, let us now define 1 as a natural number $-$ it is the only number we know at this point.

Then, let us define addition according to the a source that Wikipedia quotes:

Let $n^+$ be the successor of $n$, that is, the number following $n$ in the natural numbers, so 0$^+$ = 1, 1$^+$ = 2. Define $a$ + 0 = $a$. Define the general sum recursively by $a + (b^+) = (a + b)^+$.

If we set $a$ = 1 and $b$ = 0 in that last sentence above, we get the next natural number in the sequence, 2:

$$ \begin{align}1 + 0^+ & = (1 + 0)^+ \\ 1 + 1 & = 1^+ \\ 1 + 1 & = 2 \end{align} $$

Essentially, we added 1 (in the form of 0$^+$) to 1 in order to generate 2. We could continue this to generate 3, then 4, 5, and so on. For instance, if we set $a$ = 2 and $b$ = 0, we add one more to 2 in order to get 3:

$$ \begin{align}2 + 0^+ & = (2 + 0)^+ \\ 2 + 1 & = 2^+ \\ 2 + 1 & = 3 \end{align} $$

Summing 1 $n$ times sequentially is therefore completely equivalent to generating the $n$th natural number $-$ i.e., $n$ itself, by adding 1 to the previous natural number. This sequential addition of ones could be considered the very prototype of the Telescoping Property.