# Alternating series question

1. Mar 18, 2015

### joshmccraney

Hi PF!

The other day I was showing convergence for an alternating series, let's call it $\sum (-1)^n b_n$. I showed that $\lim_{n \to \infty} b_n = 0$ and that $b_n$ was monotonically decreasing; hence the series converges by the alternating series test. but I needed also to show it did not converge to zero. the argument I used was that since $|b_1 - b_2| >0$ and that since $b_n$ monotonically decreases, we then know $\sum (-1)^n b_n > |b_1 - b_2|$. Is my intuition correct here? If so, is it ever possible to have a series described above converge to zero?

Let me know what you think!

2. Mar 18, 2015

### wabbit

First thought this was wrong, but yes.

Set $a_n=b_{2n}-b_{2n+1}\geq 0$

$\sum(-1)^n b_n =\sum a_n \geq 0$ and this can only be zero if $\forall n, b_n=0$

Edit you seem to have the index n starting at 1 instead of 0 so you need to adjust the above a little, but this doesn't change anything except perhaps a sign.
And your argument, assuming the b's are strictly decreasing, works - you can also complete it to cover all cases where b is not identically 0.

Last edited: Mar 19, 2015
3. Mar 19, 2015

### joshmccraney

Thanks! I thought so but wanted reassurance.

4. Mar 19, 2015

### mathman

Wrong conclusion! $a_n=0$ will hold if pairs of alternate terms have the same magnitude $b_{2n}=-b_{2n+1}$, and $b_{2n} \gt b_{2n+2}$.

5. Mar 19, 2015

### joshmccraney

But $b_n$ is monotonically decreasing.

6. Mar 19, 2015

### wabbit

Oops you are right of course !

The sum is 0 iff $\forall n, b_{2n}=b_{2n+1}$ and that's all we can say.

Of course if we know that b is strictly dereasing this cannot happen, but i was trying to avoid needing that since op did not say "strict".

Failed attempt, sorry.

To expand, the generic counterexampe to my initial claim is the altermating sum
$b_0-b_0+...+b_n-b_n+...$
which converges to 0 iff $b_n\rightarrow 0$

Last edited: Mar 19, 2015
7. Mar 22, 2015

### LAZYANGEL

I'd use comparison test with another series that has similar rate of change as the one you're using.

$\sum \limits_{i} \left | {b_i} \right | \geq \sum \limits_{i} \left | {a_i} \right |$

Where $a_i \approx b_i$ in structure, but $a_i$ is both monotonic and bounded.