$(u_n)_{n\in\Bbb{N}}$ is a sequence in $\Bbb{R}$ and $u_n \to 0$,

$$u_{n+1} \le u_n, \ \forall n \in \Bbb{N} \\ u_n \ge 0, \ \forall n \in \Bbb{N}.$$

Define $s_n = \sum_{k=1}^{n}(-1)^{k}u_k \ .$

Prove that $\sum_{k=1}^{\infty}(-1)^{k}u_k$ converges, thus $s_n = \sum_{k=1}^{n}(-1)^{k}u_k$ converges.

Hints: Define $r_n = s_{2n+1}$ and $q_n = s_{2n}$. Show that $r_n$ converges, and $q_n$ has the same limit converges.

I did so:

We claim there exists $ A\in\Bbb{R},$ so that $A\ge0,$ is the sum of $\sum_{k=1}^{\infty}(-1)^{k}u_k$.

There exists an $\epsilon$ in $\Bbb{N}$, there is a $N_{\epsilon} \in \Bbb{N}$ and for all $n\ge N_{\epsilon}\\$ is $$|\sum_{k=1}^{\infty}(-1)^{k}u_k - A| < |s_{2n+1} - A|< |s_{2n} - A| < |s_n - A| < \epsilon. \\\\$$

It's clear that $$ \ r_n = q_n - u_{2n+1} \\ $$ and $$-u_{2n+1}<0<\frac{\epsilon}{2}<\epsilon.$$

Then $$|r_n - A|=|q_n - u_{2n+1} -A| \le |q_n - A|+u_{2n+1}< \epsilon \\$$

so that $$|q_n -A|< \epsilon - u_{2n+1}< \frac{\epsilon}{2} < \epsilon.$$

This means $q_n$ and $r_n$ have the same limit and are monotonically decreasing, therefore $\sum_{k=1}^{\infty}(-1)^{k}u_k$ converges to 0.

Did I right? If not, could you give me any hints to understand what is wrong, please? Thanks in advance.

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    $\begingroup$ I haven't read your proof in details but please check the wiki proof of Alternating series test $\endgroup$ – sciona Jan 15 '15 at 18:14
  • $\begingroup$ @sciona I edited the post. And I also didn't read the wikipedia page because inside there is the answer of this problem. $\endgroup$ – MadMathFourierRoad Jan 15 '15 at 19:50
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    $\begingroup$ The title is now misleading. The question is to show that a series converges, not that it converges to $0$. $\endgroup$ – robjohn Jan 15 '15 at 22:30

$s_n = -u_1 + u_2 \mp ... \mp u_{n-1} \mp u_n$

Better to use $+$ or $-$ except at the end. And for $q_n$ and $r_n$ you know the sign of the last term too.


You haven't really made a complete thought here. What are the previous few lines supposed to tell you about this sum?

Hypothesis: $ \exists A\in\Bbb{R}, \ A\ge0, r_n \to A$

In formal writeups, use English words to make your work more readable. Symbols are better for board work and scratch paper. So you might write this instead as

We claim there exists $A \in \mathbb{R}$ such that $A \geq 0$ and $r_n \to A$.

But you still haven't shown that.

$\Rightarrow \exists\epsilon\in\Bbb{N}, \ \text{for $\forall n\ge N_{\epsilon}$ is $|r_n - A| < \epsilon$} \\ $.

This doesn't make sense either. What I think you mean to say is:

(Therefore?) For every $\epsilon > 0$ there exists $N_\epsilon \in \mathbb{N}$ such that $|r_n - A| < \epsilon$ for all $n \geq N_\epsilon$.

This would be true if $r_n \to A$, which you still haven't shown.

Since $ \ r_n = q_n - u_{2n-1} \ $ follows that $|r_n - A|=|q_n - u_{2n+1} -A| \le |q_n - A|+u_{2n+1}< \epsilon $

You are trying to make the claim that $q_n \to A$ as well, so really you should start with $|q_n - A|$ on the left and use $|r_n - A| < \epsilon$ on the right.

You also need to use that $u_{2n+1} < \epsilon$ for $n$ sufficiently large. And then the sum would be $2\epsilon$. So you could go back to the beginning and replace all the $\epsilon$s with $\frac{\epsilon}{2}$, or just not worry about it. It depends on how rigorous your instructor wants you to be.

It means that $ \ q_n \to 0 \ \text{since $A \ge 0$}$,

If $q_n\to0$ then the series converges to zero, and I don't think that's what you're trying to show.

and so the hypothesis is demonstarted.

Again, you haven't said anything demonstrating whether $\{r_n\}$ converges. What you have shown is (close to) a proof that if $r_n \to A$, then $q_n \to A$ as well. This is true and part of the proof that the series converges.

Here is a hint for the part you are missing: is $\{q_n\}$ or $\{r_n\}$ monotone?

  • $\begingroup$ Thank you you was helpful. I think I understood the errors and edited my demonstration. Am I stil wrong? $\endgroup$ – MadMathFourierRoad Jan 15 '15 at 19:48
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    $\begingroup$ You still haven't addressed whether $\{r_n\}$ converges. I think it would be best if you sought out the help of your instructor. StackExchange is good for some things, but maybe not extended tutoring. $\endgroup$ – Matthew Leingang Jan 15 '15 at 20:41

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