Bounded alternating series

**chiph588@** · May 19th 2010, 05:50 PM

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

**Bruno J.** · May 19th 2010, 05:53 PM

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

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

**chiph588@** · May 19th 2010, 06:06 PM

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?

**Bruno J.** · May 19th 2010, 06:48 PM

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.

**ques** · May 19th 2010, 10:19 PM

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.

**simplependulum** · May 20th 2010, 01:37 AM

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

**ques** · May 20th 2010, 02:09 AM

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.

**chiph588@** · May 20th 2010, 05:29 AM

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.

**Bruno J.** · May 20th 2010, 07:06 AM

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.

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