1. ## Bounded alternating series

Does $\{a_n\}$ converging imply $\sum_{n=1}^\infty (-1)^na_n$ is bounded?

2. No, take $a_n=\frac{(-1)^n}{n}$.

It's true if $a_n>0$ though.

3. Originally Posted by Bruno J.
No, take $a_n=\frac{(-1)^n}{n}$.

It's true if $a_n>0$ though.
Whoops, I meant to say $a_n>0$. Why is this true?

4. Originally Posted by chiph588@
Whoops, I meant to say $a_n>0$. Why is this true?
I'm sorry, it's false! Take $a_n=1+\frac{(-1)^n}{n}$...

It's true if $a_n>0$ and converges monotonically.

5. Originally Posted by Bruno J.
I'm sorry, it's false! Take $a_n=1+\frac{(-1)^n}{n}$...

It's true if $a_n>0$ and converges monotonically.

Take $a_n=2 - 1/n$

1/(n+1) < 1/n
=> 2 - 1/(n+1) > 2 - 1/n

Here $a_n>0$ , is strictly monotone converging to 2.

So, $lim|a_n|=2$ and not 0. For a convergent series we need that the tail end of the series go to 0.

6. So it should be false , but can we find all the sequences that are the counterexamples for this statement ?

7. I just gave you one such example.

For any C > 0 ,

take $a_n=C-(1/n)$

$(a_n)$ converges to C , however $\sum_{n=1}^\infty(-1)^na_n$ is not convergent.

8. Originally Posted by ques
I just gave you one such example.

For any C > 0 ,

take $a_n=C-(1/n)$

$(a_n)$ converges to C , however $\sum_{n=1}^\infty(-1)^na_n$ is not convergent.
I was asking whether the sum is bounded, not convergent.

9. Originally Posted by ques
I just gave you one such example.

For any C > 0 ,

take $a_n=C-(1/n)$

$(a_n)$ converges to C , however $\sum_{n=1}^\infty(-1)^na_n$ is not convergent.
What chip said! The partial sums you give are bounded.

To prove it, let $b_n=a_n-a$ where $a=\lim_{n\to \infty}a_n$. By the alternating series test, $\sum_{j=1}^\infty (-1)^jb_j$ converges, which immediately implies the partial sums $\sum_{j=1}^m(-1)^ja_j$ are bounded.