The author of this article (click here for image), when proving the alternating series test, computes the limit of the odd sequence as follows:



How is $s_{2n+1}=s_{2n}+b_{2n+1}$?

Thanks in advance.


The definition of the notation $s_n$ at that reference is $$s_n=b_1-b_2+b_3+\cdots+(-1)^{n-1}b_n$$ Therefore $$\begin{align*} s_{2n}&=b_1-b_2+b_3+\cdots-b_{2n}\\ s_{2n+1}&=b_1-b_2+b_3+\cdots-b_{2n}+b_{2n+1} \end{align*}$$ so that $$s_{2n+1}=s_{2n}+b_{2n+1}$$

  • $\begingroup$ Isn't $ s_{2n}=s_{2n−2} +b_{2n-1}-b_{2n}$? $\endgroup$ – Jane Smith Jul 11 '15 at 4:43

By definition, $s_{k}$ is the sum of the first $k$ terms in the sequence: $$ s_{2n+1} = \underbrace{ \ b_1 - b_2 + \dots + (-1)^{2n+1} b_{2n} \ }_{= \ s_{2n}} + \underbrace{ \ (-1)^{2n+2} b_{2n+1} \ }_{= \ b_{2n+1}} $$


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