Help with these complex analysis series problems?

[nontradstuden]

I don't even know where to start or go with this first problem.

A) Assume that 'a sub n' E C and consider rearrangements of the convergent series the 'sum of 'a sub n' from n=1 to infinity'. Show that each of the following situations is possible and that this list includes all possibilities.

1) The sum of a sub n from n=1 to infinity converages absolutely and hence all rearrangements converge to the same value.

2) The set of possible values of convergent rearrangements is all of C.

3) The set of possible values of convergent rearrangements is an arbitrary line in C.

B) Show that the sum of [ (-1)^(n+1) / n ] from n=1 to infinity converges conditionally. The value of the sum is log(2), approcimately .69. Show that by grouping the terms by taking two positive terms, then one negative term, then two positive terms, then one negative term, and so on, the series adds up to a number larger than 1. The value of the sum therefore depends on the order in which the terms are summed.

This is all that I could come up with:

The abs value of the series diverges because it's the harmonic series where p=1.
The series itself converges by the alternating series test because the nth term converges to zero and the series is decreasing. Therefore the given series converges conditionally.

Now for the rearranging:

1 + 1/3 -1/2 + 1/5 + 1/7 -1/4 and so forth. I then just listed the partial sums.

s1= 1 + 1/3 -1/2= 5/6
s2= s1 + 1/5 + 1/7 -1/4= 0.926
s3= s2 + 1/13 + 1/15 - 1/8= 0.98013
and so on...

I know that this isn't right. I don't know how I'm supposed to show that the sum of the rearranged series is greater than 1.

C) Express the rearranged series *found above in B* in the form 'the sum of b sub n from n=1 to infinity', and find a simple expression for b sub n. Try to find the infinite sum.

After a lot of erasing I came up with:


1 + 1/3 -1/2 + 1/5 + 1/7 -1/4 and so forth can be expressed as

the sum of 1/ (4k-3) + 1/(4k-1) - 1/(2k) from k=1 to infinity. I don't know if this is correct for 'b sub n' form or how to find the infinite sum of this rearranged series.

D) Assume that 'a sub n' E R(real) and that the sum of 'a sub n' from n=1 to infinity converged conditionally. Fix an arbitrary L(limit) E R(real). Show that we can rearrange the terms to make the sum converge to L.
I don't know where to start or where to go with this one, too. Do I treat 'a sub n' as an alternating series that converges conditionally?

I know that you aren't here to solve it for me, but any help you can give me would be great. Thank you very much.

Please, let me know if I need to clarify anything.

 

[kru_]

For part A, unless you were given a specific sequence a_n to look at, the problem is asking you to use the definition of a convergent series to show that any convergent series can be rearranged to converge to any value you want. I think this is similar to the http://mathworld.wolfram.com/RiemannSeriesTheorem.html" .

For part B, you are on the right track. You'll need to calculate out to s7 to show it is > 1.

For C, try turning your new formula into a single term and see if you can sum it.

The way I think about D is that, for any convergent alternating series, we have an infinite number of positive terms and an infinite number of negative terms. Consider the alternating harmonic series. Let's say we want to make it equal to 100. We can take all of the positive (odd) terms that we want, and add them together to get to eventually reach 100. Then we add -1/2. Then we add more positive terms to get back to 100. Then we add -1/4. Then we add more positive terms to get back to 100, and so on. Since we never run out of positive terms, we can keep going with this insane ratio of positive terms to negative terms forever. Hence, the series converges to any limit that we want.

 

[nontradstuden]

@Kru.

Thanks!

Your response helped a lot.
I will rework the problems. This is my first analysis course, so it is tough for me.


Edit:

I just wanted to say that I asked my professor about these problems today and he said that he didn't know they were this difficult.

He said that 'B' was more difficult/ time consuming than he originally thought. I feel good about the little problem solving that I did do. It took a while just for me to come up with that formula. I am going to finish it up now.