First note how the notation works in the statement of the Alternating Series Test. The symbol `a_k` stands for the absolute value of the `k`th term. The term itself is `b_k=text[(]-1text[)]^(k+1)a_k`. The first bulleted condition ensures that the terms really alternate in sign; that is, there are no zero terms counted among the numbered terms. The second condition ensures that each "jump" in the sum — up or down — is smaller than the one before. And the third condition ensures that the sizes of the jumps decrease to zero; that is, the series doesn't fail to converge because of the Divergence Test. These are precisely the properties of the alternating harmonic series and the Leibniz series that we used in Section 10.5 to establish their convergence.
Specifically, we find that odd-numbered partial sums decrease and even-numbered ones increase as `n` increases: To get from `s_n` to `s_(n+2)`, we first add `b_(n+1)=text[(]-1text[)]^n a_(n+1)` and then add `b_(n+2)=text[(]-1text[)]^(n+1) a_(n+2)`. If `n` is odd, then `a_(n+1)` is being subtracted to get to `s_(n+1)` and `a_(n+2)` is being added to get to `s_(n+2)`. The amount being added is smaller than the amount being subtracted, so the next odd-numbered partial sum must be smaller. The argument for even-numbered sums is similar.
In general, the pattern for consecutive sums is to jump down on even-numbered
steps and back up on odd-numbered steps, with each jump being smaller than
the one before it. In fact, the size of the jump at step is
`text[|] s_(n+1)-s_n text[|] = text[|] b_(n+1) text[|] =a_(n+1)`, which
we know approaches `0` as `n
rarr oo`. This shows that the even-numbered and odd-numbered sums steadily
approach each other, so the sequence of partial sums must actually squeeze
down on a limiting value `S`. Furthermore, because
the sequence of even-numbered sums increases, `S` must
be bigger than every even-numbered sum. Similarly, `S` must
be smaller than every odd-numbered sum. At step `n`,
if `n` is odd, then `s_(n+1)<S<s_n`,
so `text[|] S-s_n text[|] < text[|] s_(n+1)-s_n text[|] =a_(n+1)`. At an even-numbered step, `s_n<S<s_(n+1)`, but
the conclusion is the same.