If $s_n$ is the $n^{th}$ partial sum of the alternating series , and if $s$ denotes the sum of this series, show that $|s - s_n|<z_{n+1}$
I don't know how to approach this question, I was thinking about utilizing the cauchy criterion for convergence so here is a bit of what I have done:
Proof:
Since $s$ is the sum of the series, this implies, that the series and therefore $(s_n)$, the serquence of its partial sums is convergent to $s$
Then, for all $\epsilon>0$, $\exists N \in \mathbb{N}$ such that:
$|s_m - s_n|<\epsilon$ $\forall$ $m>n>N$
In particular, choose $m=n+1$
Then, $|s_m -s_n| = |s_{n+1}-s_n| = |(-1)^{n+1}z_{n+1}|=z_{n+1} < \epsilon$
I don't know how to proceed further or even if this is the right way to approach this problem.
Can anyone please guide? I'd prefer hints at first so I can utilise them to induce a certain thought-process to get to the answer.
Thank you.
EDIT: A SECOND ATTEMPT AT THE PROBLEM:
Note: we have that $z_n \geq 0$ for all $n$ and that $z_n$ is a decreasing sequence that converges to 0.
Proof:
Noting the partial sums of the series $\sum{(-1)^{n}z_n}$, we have that:
$s_1 = -z_1$
$s_2 = -z_1 +z_2$
$s_3 = -z_1 +z_2 -z_3$
.
.
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$s_{2n}= \underbrace{-z_1 +z_2}_\text{<0} -z_3 +.......+\underbrace{-z_{2n-1} +z_{2n}}_\text{< 0}$
$s_{2n+1}= -z_1 +\underbrace{z_2 -z_3}_\text{>0} +.......+\underbrace{-z_{2n-1} +z_{2n}-z_{2n+1}}_\text{>0}$
We then have that $s_{2n}$ is decreasing and $s_{2n+1}$ is increasing and that:
$s_{2n+1} \leq s_{2n}$
Also, $s_{2n+1}$ and $s_{2n}$ will converge to the same limit, say $s$ that the series converges to by the alternating series test. This implies:
$s_{2n+1} \leq s \leq s_{2n}$
From here, it is clear that
$|s-s_{2n}| \leq |s_{2n+1} - s_{2n}|=|z_{n+1}|$
Questions: Is this proof correct and would it be ok to leave it at this? Also, can we go from here and derive $|s - s_n|<z_{n+1}$? If yes, then how?
To prove the Cauchy's criterion you have shown that the sequence $\left(s_{2n}\right)_{n \in \mathbb{N}}$ is decreasing and $\left(s_{2n+1}\right)_{n \in \mathbb{N}}$ is increasing with $$ s_{2n+1} \leq s \leq s_{2n} $$ Hence you have $$ \left|s-s_{2n}\right|=\left|R_{2n}\right|=s_{2n}-s \leq z_{2n+1} $$ and $$ \left|s-s_{2n+1}\right|=\left|R_{2n+1}\right|=s-s_{2n+1} \leq z_{2n+2} $$ Here is your result.