Abel’s Lemma,(adsbygoogle = window.adsbygoogle || []).push({});

Let [itex]a_0,a_1,a_2,\cdots [/itex] and [itex]b_0,b_1,b_2,\cdots [/itex] be elements of a field;

let[itex] s_k = a_0 + a_1 + a_2 + \cdots + a_k [/itex] k= 0,1,2,… And s_{-1}=0.

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

[tex] \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}[/tex]

Proof:

Expanding the terms of the sum gives

[tex] (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}[/tex]

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:

[tex] (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}[/tex]

[tex] 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}[/tex]

Now group canceling terms:

[tex] 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 [/tex]

All that remains is

[tex] \sum _{k=m} ^ n a_k b_k [/tex]

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

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# Abel's Lemma

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