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

Prove that $$\sum_{i = m}^n a_i + \sum_{i = n + 1}^p a_i = \sum_{i = m}^p a_i$$, where $$m ≤ n are integers, and $$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 $$n \ge m - 1 \sum_{m}^{n+1}a_i = \sum_{m}^{n}a_i + a_{n+1}$$

• What are your axioms about summation? – marty cohen Nov 25 '19 at 20:05

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

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

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

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 $$n+k=p$$.

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.