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:


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.


Note: we have that $z_n \geq 0$ for all $n$ and that $z_n$ is a decreasing sequence that converges to 0.


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?

  • $\begingroup$ What is $z_n$, the general term of the series ? $\endgroup$
    – Atmos
    Apr 1 '18 at 14:34
  • $\begingroup$ That is not given in the question. The question, as I've posted it, is complete $\endgroup$
    – Alea
    Apr 1 '18 at 14:34
  • $\begingroup$ 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 $ ) $\endgroup$
    – Atmos
    Apr 1 '18 at 14:36
  • $\begingroup$ 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 $2n+1$). $\endgroup$ Apr 21 '18 at 8:23
  • 1
    $\begingroup$ 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 $m$. $\endgroup$ 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.

  • $\begingroup$ 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? $\endgroup$
    – Alea
    Apr 1 '18 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.