Proving $ \sum\limits^{n}_{i=m}a_i+\sum\limits^{p}_{i=n+1}a_i=\sum\limits^{p}_{i=m}a_i $

Question

Prove that :

Let $m\leq n<p$ be integers and let $a_i$ be a real number assigned to each integer $m\leq i\leq p$. Then we have $$\sum\limits^{n}_{i=m}a_i+\sum\limits^{p}_{i=n+1}a_i=\sum\limits^{p}_{i=m}a_i $$

I am following analysis 1 by Tao. He defines series recursively. So we cannot use $$\sum\limits^{n}_{i=m}a_i = a_m+a_{m+1} \cdots+a_n$$ Here we should use (according to me) induction or backward induction. But in proof by induction I am stuck at the following problem regarding base case $$a_m+ \sum\limits^{p}_{i=m+1}a_i $$.
I had similar problem when I tried backward induction.
How to get past this ?

Answer 1

You're doing your induction from the wrong end. First prove that $\sum_{i=m}^na_i + \sum_{i=n+1}^{n+1}a_i = \sum_{i=m}^{n+1}a_i$ (this should be straightforward from your recursive definition). Then, supposing that $\sum_{i=m}^na_i + \sum_{i=n+1}^pa_i = \sum_{i=m}^pa_i$, show that $\sum_{i=m}^na_i + \sum_{i=n+1}^{p+1}a_i = \sum_{i=m}^{p+1}a_i$. For that, use your recursive definition for $\sum$ on the second summation, not the first.

Answer 2

Write:

$$\sum\limits_{i=m}^p a_i = a_m+a_{m+1}+ \cdots+a_p$$

Then place an imaginary vertical bar through any plus sign to get your result.

Answer 3

I'm going to add this because its cute;

$$\sum\limits^{n}_{i=m}a_i+\sum\limits^{p}_{i=n+1}a_i=\sum\limits^{p}_{i=m}a_i$$ $$\sum\limits^{n}_{i=m}a_i+a_{n+1}+\sum\limits^{p}_{i=n+2}a_i=\sum\limits^{p}_{i=m}a_i$$ $$\sum\limits^{n+1}_{i=m}a_i+\sum\limits^{p}_{i=n+2}a_i=\sum\limits^{p}_{i=m}a_i$$