I had asked the same question here https://matheducators.stackexchange.com/questions/9999/which-concept-did-i-overlook-in-trying-to-prove-a-property-of-finite-series-sum but it was suggested as off-topic and hence I am asking the it here again.
In the book that I am using a finite series is defined as follows
Definition (Finite series). Let $m,n$ be integers, and let $(a_i)_{i=m}^{n}$ be a finite sequence of real numbers, assigning a real nmber $a_i$ to each integer $i$ between $m$ and $n$ inclusive. Then we define the finite sum $\sum_{i=m}^{n} a_i$ by the recursive formula $\sum_{i=m}^{n} a_i = 0 $ whenever $n < m$ and $\sum_{i=m}^{n+1} a_i = \sum_{i=m}^{n} a_i + a_{n+1}$ whenever $ n \geq m-1$ .
Question In The Book Let $m \leq n$ be integers, $k$ be another integer. Then $\sum_{i=m}^{n} a_i = \sum_{j=m+k}^{n+k} a_{j-k}$. Prove by induction.
My attempted solution For $k=0$ we get $\sum_{i=m}^{n} a_i = \sum_{j=m}^{n} a_{j}$ which is true. Assume inductively that the relation holds for any $k$. So for $k+ 1$ we have $\sum_{j=m+k+1}^{n+k+1} a_{j-(k+1)}$. However after this step I am not able to proceed further i.e., I am not able to see how to reduce this expression to $\sum_{i=m}^{n} a_i$.
MY QUESTION
- Where do I lack in my understanding? Do I lack in my understanding of induction, the definition of finite series or some simple algebraic manipulation?
Note that $$\begin{aligned}\sum_{j=m+k+1}^{n+k+1}a_{j-(k+1)}&=\sum_{j=\color{red}{(m+1)}+k}^{\color{red}{(n+1)}+k}a_{(j-1)-k}\\&=\sum_{i=m+1}^{n+1}a_{i-1}\qquad{\text{by induction hyphotesis}}\end{aligned}$$
Because by hypothesis, this is true for all natural numbers $n,m$, and we have that $n+1$ and $m+1$ are natural numbers.
Now, $\sum_{i=m+1}^{n+1}a_{i-1}=\sum_{i=m}^na_i$ is the base case. Thus, $$\sum_{j=m+k+1}^{n+k+1}a_{j-(k+1)}=\sum_{i=m}^na_i.$$
By other hand, we cannot prove all the claim by induction becuase $k$ is a integer (such as $n$). It is necessary prove when $k$ is a negative integer. The induction only works to prove "for every natural number", it does not "for every integer".
But we know that $$m\le n\quad\text{iff}\quad n=m+d\text{ for some natural number } d.$$ (Note that this is true when we consider $0$ a natural number.)
Now, using this fact, we can induct on $d$ and show the claim:
Clearly, the base case $d=0$ is true by $$\sum_{i=m}^na_i=\sum_{i=m}^{m+0}a_i=a_m=a_{m+k-k}=\sum_{j=m+k}^{m+0+k}a_{j-k}=\sum_{j=m+k}^{n+k}a_{j-k}.$$ Now, using the induction hyptohesis to prove when $d+1$, we have $$\begin{aligned}\sum_{j=m+k}^{n+1+k}a_{j-k}&=\sum_{j=m+k}^{m+d+1+k}a_{j-k}\\&=\left(\sum_{j=m+k}^{m+d+k}a_{j-k}\right)+a_{m+d+1+k-k}\\&=\left(\sum_{i=m}^{m+d}a_i\right)+a_{m+d+1}\\&=\sum_{i=m}^{m+d+1}a_i\\&=\sum_{i=m}^{n+1}a_i\end{aligned}$$
EXTRA. We need to consider the definition of sequence (in this case, we have two sequences $(a_{n+1})_{n=m}^\infty$ and $(a_n)_{n=m+1}^\infty$ ). Technically, the sequences $(a_{n+k})_{n=m}^\infty$ and $(a_n)_{n=m+k}^\infty$ are different sequences, because they are defined on different domains. However they are indeed very closely related by carefully expanding out all the definitions.
Definition of sequence. Let $m$ be an integer. A sequence $(a_n)_{n=m}^\infty$ of real numbers is any function $f\colon\{n\in\mathbf Z:n\ge m\}\to\mathbf R$.