Question

How do you determine whether a series is absolutely convergent or conditionally convergent?
It would be nice if you could give detailed solution or at least video resources from MIT.
Following is the problem.
http://dl.dropbox.com/u/63664351/Calculus%20-%20Maclaurin's%20series%20-%20concept%20of%20'convergence'%20and%20approximation'.PNG

Answer 1

Let say problem a) for example. I know it is convergent, since \[a_k=0 \] as k -> infinity.
But how do I determine whether it is conditionally convergent?

Answer 2

a) is absolutely convergent if

\[\sum_{k=3}^\infty \frac{1}{\log(k)} \]
converges. You can compare this to the harmonic series
\[\sum_{k=3}^\infty \frac{1}{k} \]

since for all k, log(k) < k, it follows that 1/log(k) > 1/k. Because the harmonic series diverges, it is clear that our series is divergent as well. That means that the original series in question converges but does not converge absolutely, i.e. it is conditionally convergent.

Answer 3

I see, so for c) does that mean it is absolutely convergent since you can't compare it with 1/k ?

Answer 4

How about d)?

Answer 5

The comparison is a means by which you can determine if a series converges or diverges. A series
\[\sum a_n\] is called absolutely convergent if
\[\sum |a_n| \]
converges as well.

Answer 6

To answer your question without solving the problem, which you should be able to do, c) is absolutely convergent and d) is not.

Answer 7

I can do c) but I can't do d). It would be glad if you can provide me with your solution on d).
Thanks.

Answer 8

Taking the absolute value of the terms of d), the asymptotic behavior of the summand is of the form
\[\sum \frac{1}{\sqrt{k}} = \sum \frac{1}{k^{1/2}} \]
We know that if you have a sum of the form
\[\sum \frac{1}{k^p} \]
then it converges if p > 1 and diverges otherwise. Thus, the above series diverges.

Answer 9

Nice one!