Proof Help: (Series) Bartle Exercise 9.3 Q 4

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:

$s_1 = -z_1$

$s_2 = -z_1 +z_2$

$s_3 = -z_1 +z_2 -z_3$

.

.

.

$s_{2n}= \underbrace{-z_1 +z_2}_\text{<0} -z_3 +.......+\underbrace{-z_{2n-1} +z_{2n}}_\text{< 0}$

$s_{2n+1}= -z_1 +\underbrace{z_2 -z_3}_\text{>0} +.......+\underbrace{-z_{2n-1} +z_{2n}-z_{2n+1}}_\text{>0}$

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?

• What is $z_n$, the general term of the series ? Apr 1 '18 at 14:34
• That is not given in the question. The question, as I've posted it, is complete
– Alea
Apr 1 '18 at 14:34
• Ok so to clarify it, you got a series $\displaystyle \sum_{n \geq 0} \left(-1\right)^n z_n$, and you need to show that $\left|R_n\right| \leq z_n$ which is a well-known equality. ( i clarify this because I dudce that you did not know what was $z_n$ ) Apr 1 '18 at 14:36
• This proof given now is correct, only at the end you have $z_{2n+1}$ instead of $z_{n} + 1$. To complete the proof for $|s-s_n| < z_{n+1}$, you can do it for $n$ even and $n$ odd, both of which follow from the statements of $2n$ and $2n+1$ (for example, you derived for $2n$ above, but same thing works out for $2n+1$). Apr 21 '18 at 8:23
• See, you proved that $|s - s_{2n}| \leq z_{2n+1}$ above. I am saying, that similarly you can prove that $|s - s_{2n+1}| \leq z_{2n+2}$. Once you do this, suppose you want to prove that $|s-s_{m}| \leq z_{m+1}$ for any $m$. If $m$ is odd, then you can use the second statement above, and if $m$ is even then you can use the first statement of the comment. Therefore, you have proved $|s-s_m| \leq z_{m+1}$ for all $m$. Apr 21 '18 at 23:28

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.
• How did you deduce that $(s_{2n+1})$ is decreasing whilst $(s_{2n})$ is increasing? Reminds me of how we go about proving convergence of alternating series showing that both these sequences converge to the same value. However, I did not know that it will always be the case for a convergent alternating series to have$(s_{2n+1})$ decreasing whilst $(s_{2n})$ increasing. Can you elaborate a bit on that?