Proving Convergence of Complex Series with Cauchy Condition | Homework Help

In summary, the homework equation for sums of complex numbers is that the sum converges for every sequence of complex numbers with the property that the sum of the largest n terms is bounded below by the harmonic series.
  • #1
gustav1139
14
0

Homework Statement



Suppose that [itex]\left\{a_{n}\right\}[/itex] is a sequence of complex numbers with the property that [itex]\sum{a_{n}b_{n}}[/itex] converges for
every complex sequence [itex]\left\{b_{n}\right\}[/itex] such that [itex]\lim{b_{n}}=0[/itex]. Prove that [itex]\sum{|a_{n}|}<\infty[/itex].

Homework Equations


The Attempt at a Solution



I tried going directly, at it, using the Cauchy condition on [itex]\sum{a_{n}b_{n}}[/itex] to try to figure something out about the [itex]a_{n}[/itex], but I got bogged down, and all the inequalities seemed to be pointing the wrong direction.
Then I tried to prove the contrapositive, that if [itex]\sum{|a_{n}|}[/itex] diverges, then there exists a sequence [itex]\left\{b_{n}\right\}[/itex] such that [itex]\sum{a_{n}b_{n}}[/itex] diverges as well. But I didn't get very far with that either.
I was able to prove it using the ratio test (i.e. if [itex]\lim{\frac{a_{n+1}b_{n+1}}{a_{n}b_{n}}}=r[/itex] where [itex]r\in(0,1)[/itex]), but that's unfortunately not a necessary condition for convergence.
So I'm stuck, and frustrated :(
 
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  • #2
What you need to prove is that if [itex]\displaystyle \int_{0}^{\infty} a(x)b(x)\,dx[/itex] has a finite value, so must [itex]\displaystyle \int_{0}^{\infty} |a(x)|\,dx[/itex]. You can get around the absolute value assuming that a(x) resolves to a non negative real number for every x.

Try using the first mean value theorem for integration.
 
  • #3
Thanks! I'll give that a try, but isn't it an issue that they're complex sequences, if I want to use the MVT?

Also, why can you assume a(x) is a nonnegative real? Or perhaps I'm misunderstanding what you mean by resolve.
 
  • #4
gustav1139 said:
Then I tried to prove the contrapositive, that if [itex]\sum{|a_{n}|}[/itex] diverges, then [itex]\sum{a_{n}b_{n}}[/itex] must diverge as well. But I didn't get very far with that either.
You didn't get the contrapositive quite right. Consider, for example, an=1 and bn=1/n2. [itex]\sum{|a_{n}|}[/itex] diverges but [itex]\sum{a_{n}b_{n}}[/itex] doesn't.

You should have said: if [itex]\sum{|a_{n}|}[/itex] diverges, then there exists a sequence {bn} such that ##\displaystyle\lim_{n \to \infty} b_n = 0## and [itex]\sum{a_{n}b_{n}}[/itex] diverges.

That said, I don't have any comment about how to actually do the proof. Just wanted to point out your error.
 
  • #5
You could make a proof by contradiction. Suppose ## \sum | a_i|## diverges. Therefore you can find a sequence of integers ##k_n## such that the partial sums ## \sum_{k_{n-1}+1}^{k_n} | a_i| > 1## for n = 1, 2, 3 ...

Now choose a series of b's that converge to 0, such that ##\sum a_ib_i## diverges.

Hint: you can choose b's such that every term ##a_i b_i## is real and non-negative...
 
  • #6
vela said:
You didn't get the contrapositive quite right.

Right you are. All fixed.
That (what you said) is however what I'd been working with so that's still no go...
 
  • #7
AlephZero said:
You could make a proof by contradiction. Suppose ## \sum | a_i|## diverges. Therefore you can find a sequence of integers ##k_n## such that the partial sums ## \sum_{k_{n-1}+1}^{k_n} | a_i| > 1## for n = 1, 2, 3 ...

Now choose a series of b's that converge to 0, such that ##\sum a_ib_i## diverges.

Hint: you can choose b's such that every term ##a_i b_i## is real and non-negative...

Thanks! I think that broke it open. I had been restricting myself unnecessarily with what I could assume if ##\sum|a_i|## diverged.

I picked ##b_k = \frac{\overline{a_i}}{c_i |a_i|}## where the c's are the largest n such that i > k_n. Then the b's go to 0 since they do in modulus, and you can group the terms in the product series to give something bounded below by the harmonic series. boom.
 

Related to Proving Convergence of Complex Series with Cauchy Condition | Homework Help

1. What is a tricky complex series?

A tricky complex series is a mathematical concept that involves a sequence of complex numbers, where each term is related to the previous term by a certain pattern or rule. These series can be challenging to analyze and often require advanced techniques to determine their convergence or divergence.

2. How do you determine the convergence of a tricky complex series?

There are several tests that can be used to determine the convergence of a tricky complex series, such as the ratio test, the root test, and the comparison test. These tests involve evaluating the behavior of the series as the number of terms increases and can help determine if the series will converge or diverge.

3. What are some common examples of tricky complex series?

One common example of a tricky complex series is the Taylor series, which is used to represent a function as an infinite sum of terms. Another example is the geometric series, where each term is multiplied by a constant factor.

4. Can a tricky complex series have multiple points of convergence?

Yes, it is possible for a tricky complex series to have multiple points of convergence. This means that the series can converge to different values depending on the starting point or the conditions of the series. It is important to carefully analyze the behavior of the series to determine all possible points of convergence.

5. How are tricky complex series used in real-world applications?

Tricky complex series are used in a variety of scientific and mathematical fields, such as physics, engineering, and finance. They can be used to model and analyze complex systems, make predictions, and solve problems in these fields. For example, the Taylor series is used in physics to approximate the behavior of a system near a specific point, while the geometric series is used in finance to calculate compound interest.

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