I Proof of Alternating Series Test

[Mr Davis 97]

I'm looking at the proof of the alternating series test, and the basic idea is that the odd and even partial sums converge to the same number, and that this implies that the series converges as a whole. What I don't understand is why the even and odd partial sums converging to the same limit implies that the series converges as a whole.

 

[jedishrfu]

These may explain it:

https://math.dartmouth.edu/archive/m8w10/public_html/m8l09.pdf

https://ocw.mit.edu/ans7870/18/18.013a/textbook/HTML/chapter30/section02.html[URL]https://math.dartmouth.edu/archive/m8w10/public_html/m8l09.pdf[/URL]

 

[Mr Davis 97]

These may explain it:

https://math.dartmouth.edu/archive/m8w10/public_html/m8l09.pdf

https://ocw.mit.edu/ans7870/18/18.013a/textbook/HTML/chapter30/section02.htmlhttps://math.dartmouth.edu/archive/m8w10/public_html/m8l09.pdf

It didn't satisfactorily answer my question. Essentially, my question is this: If $S_n$ is the nth partial sum of our alternating series, why does $\lim S_{2m} = L = \lim S_{2m+1}$ imply that $\lim S_n = L$?

 

[mathwonk]

that follows immediately from the definition of a limit. I.e. roughly speaking if, in the sequence of partial sums, all elements with large enough even index as well as all those with large enough odd index are as close as you wish to L, then in fact all elements with large enough index are as close as you want to L, and that is the definition of convergence to L.

But to me the basic idea is to use the Cauchy convergence test, since the distance between any two partial sums Sn and Sm, with n<m, is at most equal to the absolute value of the mth series element am, and this goes to zero by hypothesis.