Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My textbook gives this definition of the alternating series test:

Test for alternating series. An alternating series conerges if the absolute value of the terms decreases steadily to zero, that is, if $|a_{n_1}| \le |a_n|$ and $\lim_{n\to\infty}a_n = 0$.

Similarly, Wikipedia gives this definition:

...the alternating series test tells us that an alternating series will converge if the terms $a_n$ converge to 0 monotonically.

I don't see an "if and only if" in either definition. My question is: If the terms $a_n$ do not converge to 0 monotonically, do I know that the series diverges? Or is the test inconclusive?

share|improve this question
2  
I'm not sure it would fully answer your question, but $1-0+1/2^2-0+1/3^2-0+1/4^2-0+\cdots$ converges. In general, if the series of positive terms and the series of negative terms would each converge separately, the series will converge regardless of monotonicity of the absolute values. –  Jonas Meyer Jan 28 '13 at 4:42
2  
If the terms aren't monotone, the test is inconclusive. Why not try to find examples of both types, and then post them as an answer? –  Gerry Myerson Jan 28 '13 at 4:45
    
In your question, "$|a_{n_1}| \le |a_n|$" isn't defined. Perhaps you mean $|a_{n+1}| \le |a_n|\;$ ($n=1, 2, \dots$). –  John Bentin Jan 28 '13 at 10:22
    
@JohnBentin: Yeah I noticed that, but I was copying directing from the book. Clearly what they're trying to get at is monotonicity, so I understand their intent. –  Matthew Jan 28 '13 at 18:51
add comment

1 Answer

up vote 3 down vote accepted

If the terms do not converge to $0$, then we cannot have convergence. But if monotonicity is abandoned, then we could have convergence or divergence.

For instance, the series $$-\frac{1}{2^2}+\frac{1}{1^2}-\frac{1}{4^2}+\frac{1}{3^2}-\frac{1}{6^2}+\cdots$$ converges, indeed converges absolutely.

The series $$-\frac{1}{2}+\frac{1}{1}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}+\cdots$$ also converges.

We now give an example of divergence. Write down the usual harmonic series $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$. In between all these positive terms, put in, in turn, $-\frac{1}{2}$, $-\frac{1}{2^2}$, $-\frac{1}{2^3}$, $-\frac{1}{2^4}$ and so on.

One can produce more interesting examples. Take the familiar alternating series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$.

We can rearrange the terms of this series, so that they still alternate in sign, but the resulting series diverges.

share|improve this answer
1  
Also, we can rearrange the terms of the last example (still making them alternate in sign) to make the series converge to any given real number. In fact this can be done with any series of real numbers that converges conditionally. –  Robert Israel Jan 28 '13 at 7:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.