Can I state that a of n is decreasing w/o taking the derivative

[freshman2013]

This applies to the alternating series test, and one of the conditions for it to work is if absolute value of a_n is decreasing for all n. The way the book says to do it is to take the derivative of f(x) with f(x)=a_n. However, if I know that the limit as n approaches infinity abs value a_n approaches 0, and that a_n (w/o the (-1)^n part) is positive for all n>N, shouldn't that enough to prove that a_n is decreasing to infinity. All the examples I did seem to follow this reasoning. Example: (-1)^(n-3)* sqrt(n)/(n+4). Clearly, if n is a really big positive number, then sqrt(n)/(n+4) can't be negative and the limit as it goes to infinity zero. Then the only way I can see it approaching zero is by decreasing. Might there be exceptions to this case and if so give an example? The only reason I'm asking this is that taking the derivative seems like unnecessary work to me. If I explain this on a test instead of taking a derivative, might the professor have any reason to takes points off?

 

[Office_Shredder]

Consider
a_{2n+1} = \frac{1}{(2n+1)^2}
and
a_{2n} = \frac{1}{2n}

Then
\sum (-1)^n a_n
is an alternating series, the terms are going to zero, but the series does not converge because the a_2n terms are way bigger than the a_2n+1 terms and the sequence is not decreasing term by term. It is generally true that if the a_n are given as a single rational function then if they're going to zero they will do so uniformly, but it's better not to make assumptions and just make the extra check.

 

[freshman2013]

So for series represented as one function, my reasoning should always work?

 

[gopher_p]

So for series represented as one function, my reasoning should always work?

No. $a_n=\frac{\cos^2(n\pi/2)}{n}$.

In order to use the Alternating Series Test, you must demonstrate that the sequence in question is decreasing to zero. If you don't want to do that, then you can't use the AST.

You cannot assume that a sequence of positive terms that converges to zero does so in a decreasing fashion, because that is not true in general. I know it seems like the "bad" kind of sequences must be exotic, but it's actually the "good" sequences that are rare.

In general, it can be difficult to show that a sequence is decreasing. Be grateful that you have been given such a powerful and easy-to-use tool as the derivative to help you out in some cases. It may seem like a lot of work, but I promise you it's one of the more straightforward ways of getting the job done.

I think the bottom line here is that come quiz/exam time, you should probably just do the problems using the tools that you were given in class. Keep in mind that, in addition to testing your understanding of new material, some assessments are designed to make sure you are still competent in the old material too.