# Summation Equivalent definition

Given the definition: Prove Lemma 6.1.1: However, when the proof gets to the following case, I don't understand the second "step" (marked with a red question mark). Can you provide some clarification? As fully detailed and justified as possible, please. I tried proving Lemma 6.1.1 myself using induction but I couldn't, here's my attempt:

Let $$a,b\in \mathbb{Z}$$. Note: Here induction "runs" on $$\mathbb{N}$$ starting from $$1$$ (not $$0$$):

1. $$a=b$$ \begin{align} \sum_{n=a}^{b} f(n) &= f(n)\\ &= f(b) && \text{n=a=b}\\ &= f(b) + 0\\ &= f(b) + \sum_{n=a}^{a-1} f(n) && \text{Definition (6.1.1)} \\ &= f(b) + \sum_{n=a}^{b-1} f(n) \end{align}
2. For $$a, we use induction over $$b$$,
• [Base case] For $$b=a+1$$, \begin{align*} \sum_{n=a}^{b=a+1} f(n) &= f(a) + \sum_{n=a+1}^{a+1} f(n) && \text{Definition (6.1.1)}\\ &= f(a) + \sum_{n=b}^{b} f(n)\\ &= f(a) + f(b)\\ &= f(b) + f(a)\\ &= f(b) + \sum_{n=a}^{a} f(n)\\ &= f(b) + \sum_{n=a}^{b-1} f(n) && \text{b=a+1 \Rightarrow b-1=a} \end{align*}
• [Induction hypothesis] Assume that for $$b=a+k$$, with $$k\in \mathbb{N}$$ we have that: \begin{align*} \sum_{n=a}^{a+k} f(n) &= f(b) + \sum_{n=a}^{b-1} f(n) \end{align*}
• For $$b=a+k$$ we have: \begin{align*} \sum_{n=a}^{b=a+k+1} f(n) &= f(a) + \sum_{n=a+1}^{a+k+1} f(n) && \text{Definition (6.1.1)}\\ \end{align*}

The induction in proof 6.1.1 is on $$b-a$$.

The induction hypothesis would assume the lemma is true for any LHS where the the upper and lower bounds differ by $$b-a-1$$. This includes assuming the lemma is true for LHS like:

\begin{align*} \sum_{i=a}^{b-1}f(i) &= \cdots\\ \sum_{i=a-1}^{b-2}f(i) &= \cdots\\ \sum_{i=a+1}^{b}f(i) &= \cdots \tag{*}\\ \sum_{i=1}^{b-a}f(i) &= \cdots\\ \end{align*}

Then your $$(\color{red}?)$$ line is applying one particular case $$(*)$$ of the induction hypothesis, where the LHS has lower bound $$a+1$$ and upper bound $$b$$:

$$\sum_{i=a+1}^{b}f(i) = \begin{cases} 0 & \text{if b

The summation on the RHS of the last line should have upper limits of a+k. Then you might be able to use the first term on the RHS to arrive at something that lets you use induction.