Abel's Lemma: Direct and Induction Proof

[Integral]

Abel’s Lemma,
Let $a_0,a_1,a_2,\cdots$ and $b_0,b_1,b_2,\cdots$ be elements of a field;
let$s_k = a_0 + a_1 + a_2 + \cdots + a_k$ k= 0,1,2,… And s_-1 =0.

Then for any positive real integer n and for m= 0,1,2,…,n-1,

$$\sum^n _{k=m} a_k b_k = \sum ^{n-1}_{k=m} (b_k - b_{k+1}) s_k + b_n s_n - b_m s_{m-1}$$



Proof:

Expanding the terms of the sum gives

$$(b_m - b_{m+1})s_m + (b_{m+1} - b_{m+2} )s_{m+1} + \cdots + (b_{n-1} - b_n) s_{n-1} + b_n s_n - b_m s_{m-1}$$

By the definition of s_k we have s_k+1 = s_k + a_k+1

Now for the expansion of the sum we can write:

$$(b_m s_m - b_{m+1} s_m + b_{m+1} s_{m+1} - b_{m+2} s_{m+1} + \cdots + b_{n-1} s_n - b_n s_{n-1} + b_n s_n - b_m s_{m-1}$$


$$b_m a_m + b_m s_{m-1} - b_{m+1} s_m + b_{m+1} s_m + b_{m+1} a_{m+1} - b_{m+2} s_{m+1} + \cdots + b_{n-1} s_n - b_n s_{n-1} + b_n s_n - b_m s_{m-1}$$

Now group canceling terms:

$$b_m a_m + ( b_m s_{m-1} -b_m s_{m-1}) + ( - b_{m+1} s_m + b_{m+1} s_{m}) + b_{m+1} a_{m+1} + \cdots + b_{n-1} a_{n-1} + b_{n-1} s_n - b_n s_{n-1} + b_n s_n$$

All that remains is

$$\sum _{k=m} ^ n a_k b_k$$

Is this proof sufficient or do I need do an induction proof?

 

[Galileo]

It's sufficient. But my teacher would say it's too suggestive with those dots.
Use induction. It's much neater.

 

[Integral]

OK, let me see if I can do the induction proof.
The part that has bothered me is the m, not clear to me what this buys? I guess it it a bit more generalized, don't have to start at zero.

For the induction proof I need to show that the relatationship is ture for n=1 and m=0

$$\sum _ {k=0}^1 a_k b_k = a_0 b_0 + a_1 b_1$$

for the RHS I have

$$\sum ^0 _ {k=0} (b_k - b_{k+1})s_k + b_n s_n - b_m s_{m-1}= (b_0 - b_1)s_0 + b_0 s_0 + b_1 a_0 + b_1 a_1 - b_0 s_{-1}$$
$$= b_0 a_0 - b_1 a_0 + b_0 a_0 + b_1 a_0 + b_1 a_1 - 0$$
$$= b_0 a_0 + a_1 b_1$$

So the statement is valid for n=1, now assume that the general statement for is true n= N.

Let n= N+1

$$\sum^{N+1} _{k=m} a_k b_k = \sum ^N_{k=m} (b_k - b_{k+1}) s_k + b_{N+1} s_{N+1} - b_m s_{m-1}$$
For the RHS we have

$$\sum^{N+1} _{k=m} a_k b_k = \sum ^{N-1}_{k=m} (b_k - b_{k+1}) s_k + b_{N+1} s_{N+1} - b_m s_{m-1}+( b_{N} - b_{N+1})s_N$$


$$= \sum ^{N-1}_{k=m} (b_k - b_{k+1}) s_k + b_{N+1} s_{N} + b_{N+1}a_{N+1} - b_m s_{m-1}+ b_{N}s_N - b_{N+1}s_N$$

$$= \sum ^{N-1}_{k=m} (b_k - b_{k+1}) s_k + b_Na_N - b_m s_{m-1}+ b_{N+1}a_{N+1}$$

$$= \sum ^N_{k=m} a_k b_k + a_{N+1}b_{N+1}$$

$$= \sum^{N+1}_{k+m} a_k b_k$$
QED