Alternating series where the limit goes to zero BUT it diverges?

[nlsherrill]

Homework Statement

My prof gave us an extra credit opportunity for a few extra points on the final exam(tomorrow).

He told us to go find an example of an alternating series that is decreasing, its limit->0, and it diverges. So far I haven't seen any examples, plus I have sat around some tonight instead of studying for the final trying to figure this one out.

Homework Equations

Alternating series test. But I am bad at Latex so Ill spare everyone.

The Attempt at a Solution

I'm not sure if I am right on this one, but how about an alternating p-series with p=1? If you take the limit of 1/n as n->infinity the series goes to zero. Its also decreasing AND according to the p-series test, if p<=1 then it diverges. Is this a good answer or..?

We have actually used this on before in class...so he may be looking for a more original alternating series.

By the way, I have to actually make up my own problem and solve it, but that will be the easy part.

 

[Char. Limit]

But the p-series applies to functions of the sort...

\sum_{n=1}^\infty \frac{1}{n^p}

Which is non-alternating. Also, the alternating harmonic series you describe...

\sum_{n=1}^\infty \frac{\left(-1\right)^{n-1}}{n}

Does converge, as do all alternating series that are absolutely decreasing.

 

[nlsherrill]

But the p-series applies to functions of the sort...

\sum_{n=1}^\infty \frac{1}{n^p}

Which is non-alternating. Also, the alternating harmonic series you describe...

\sum_{n=1}^\infty \frac{\left(-1\right)^{n-1}}{n}

Does converge, as do all alternating series that are absolutely decreasing.

so...no such series exist? I must have misheard my professor/

 

[Char. Limit]

so...no such series exist? I must have misheard my professor/

Or he could be cruel. I'm pretty sure no such series exists however. If one did, the alternating series test would be invalid.

 

[gomunkul51]

A Leibniz Series is:

1. alternating i.e. (-1)^n.
2. lim |a| -> 0
3. decreasing.

any series that has this 3 conditions, converges conditionally :)

 

[HallsofIvy]

Or he could be cruel.

A man after my own heart!

I'm pretty sure no such series exists however. If one did, the alternating series test would be invalid.

 

[Bohrok]

I think the trick is to find one that's decreasing to 0, but not monotonically decreasing:

1 - 0 + 1/2 - 0 + 1/3 - 0 + ···

1/1 - 1/1² + 1/2 - 1/2² + 1/3 - 1/3² + ···

 

[Char. Limit]

Maybe you could use the bernoulli numbers. Every other n, B_n is zero, so if you co-ordinated that with the (-1)^n just right...