Can an alternating series with decreasing terms converge to zero?

member 428835
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!
 
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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.
 
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Thanks! I thought so but wanted reassurance.
 
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}.
 
But ##b_n## is monotonically decreasing.
 
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 ##
 
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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.
 

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