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Prove that $\sum_{i = m}^n a_i + \sum_{i = n + 1}^p a_i = \sum_{i = m}^p a_i$, where $m ≤ n<p$ are integers, and $a_i$ is a real number assigned to each integer $m ≤ i ≤ p$. (Hint: you might want to use induction)

Let's follow the hint and use induction on $p-m = k$
Base case: $k = 1$, then $p = m + 1$ and $n = m$. $\sum_{i = m}^n a_i + \sum_{i = n + 1}^p a_i = a_m + a_{m+1}$ and $\sum_{i = m}^p a_i = a_m+a_{m+1}$. Therefore, the right-hand side is equal to the left-hand side.
Inductive step: Assume for $p-m=k$ the statement holds, show for $p-m = k + 1$. We know that $\sum_{i = m}^n a_i + \sum_{i = n + 1}^{m+k} a_i = \sum_{i = m}^{m+k} a_i$. Now, $\sum_{i = m}^n a_i + \sum_{i = n + 1}^{m+k+1} a_i = \sum_{i = m}^n a_i + \sum_{i = n + 1}^{m+k} a_i + a_{m+k+1} = \sum_{i = m}^{m+k} a_i + a_{m+k+1}$ by inductive hypothesis. Therefore, we get $$\sum_{i = m}^n a_i + \sum_{i = n + 1}^{m+k+1} a_i =\sum_{i = m}^{m+k+1} a_i$$

Is this prove plausible. At this point about the finite sum, I can use the following facts:

if $ m < n \sum_n^m a_i= 0$
if $n \ge m - 1 \sum_{m}^{n+1}a_i = \sum_{m}^{n}a_i + a_{n+1}$

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    $\begingroup$ What are your axioms about summation? $\endgroup$ – marty cohen Nov 25 '19 at 20:05
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I suppose it is convenient apply induction in a different way, that is

  • base case, $p=n+1 \implies \sum_{i = m}^n a_i + \sum_{i = n + 1}^{n+1} a_i = \sum_{i = m}^n a_i + a_{n+1}=\sum_{i = m}^{n+1} a_i$

and for the induction step assuming that $\sum_{i = m}^n a_i + \sum_{i = n + 1}^p a_i = \sum_{i = m}^p a_i$ we need to prove

$$\sum_{i = m}^n a_i + \sum_{i = n + 1}^{p+1} a_i = \sum_{i = m}^{p+1} a_i$$

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$$\sum_{i=m}^n a_i+\sum_{i=n+1}^{p}a_i=\sum_{i=m}^n a_i+a_{n+1}-a_{n+1}+\sum_{i=n+1}^{p}a_i =\sum_{i=m}^{n+1} a_i+\sum_{i=n+2}^{p}a_i$$

is a base case and by induction,$$\sum_{i=m}^n a_i+\sum_{i=n+1}^{p}a_i =\sum_{i=m}^{n+k} a_i+\sum_{i=n+k+1}^{p}a_i.$$

Form this the initial claim follows by setting $n+k=p$.

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It should be pretty straightforward.

The left-hand side is $\sum_{i = m}^n a_i + \sum_{i = n + 1}^p a_i$ which after expansion (since $m,p,n$ are finite, we can always expand explicitly) gives us $(a_m+a_{m+1}+ \cdots+ a_{n-1}+a_n)+ (a_{n+1}+ \cdots+ a_{p-1}+a_p).$

Clubbing together all the terms, we can write it as $(a_m+a_{m+1}+ \cdots a_{n-1}+a_n+a_{n+1}+ \cdots+ a_{p-1}+a_p).$ which is nothing but $ \sum_{i = m}^p a_i.$ i.e. right-hand side.

I think it qualifies as proof.

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