Convergence of Series: Homework Statement and Attempt at Solution

In summary: So we know it diverges.In summary, the conversation discusses whether the series -1/(3*k) + 1/(3*k+1) + 1/(3*k+2) converges, with the participants considering different approaches and arguments. Ultimately, it is determined that the series does not converge, using techniques such as algebra, limit comparison, and telescoping.
  • #1
ptolema
83
0
it's a new semester, and we're at it again. series abounds!

Homework Statement



Let
cond1.jpg
.
Determine whether the following series converges:
cond.jpg


Homework Equations



-1 [tex]\leq[/tex] [tex]\sigma[/tex]k [tex]\leq[/tex] 1

The Attempt at a Solution


i feel like the series inherently diverges because of the 1/k element. also,
cond2-1.jpg
, so i know that the series does not converge absolutely. of course, that isn't solid proof that the series diverges, but other convergence tests are harder to work with for this particular problem. since the [tex]\sigma[/tex]k and 1/k series both diverge, it makes sense to me that the above series would also diverge. could someone help me make this less intuitive and find a more direct approach?
 
Physics news on Phys.org
  • #2
Think about doing some algebra on -1/(3*k)+1/(3*k+1)+1/(3*k+2). Does that help you get past intuition?
 
  • #3
after a bit of algebra, I got to
frac.jpg
. I tried taking the limit as k-->infinity, and the limit turned out to be zero. from this, i can kind of tell that as k increases, the sums of consecutive terms in the series get smaller and approach zero. this suggests convergence.
also, after looking at it for a while, I decided to treat it as a weird case of alternating series. since lim 1/k as k->infinity is zero, i reached the tentative conclusion that the series converges. as any of my arguments valid?
 
  • #4
ptolema said:
after a bit of algebra, I got to
frac.jpg
. I tried taking the limit as k-->infinity, and the limit turned out to be zero. from this, i can kind of tell that as k increases, the sums of consecutive terms in the series get smaller and approach zero. this suggests convergence.
also, after looking at it for a while, I decided to treat it as a weird case of alternating series. since lim 1/k as k->infinity is zero, i reached the tentative conclusion that the series converges. as any of my arguments valid?

It's DEFINITELY not convergent. Look at the expression you got. For large k that's approximately 9k^2/(27k^3) (ignoring smaller powers). That's 3/k. You have to add those up for all k. It's like a harmonic series. To formalize it think about a limit comparison test.
 
Last edited:
  • #5
We can save ourselves some algebra and a limit comparison test. Firstly, we can ignore the first term of the series since it won't effect convergence. Call this adjusted series S. Then note the consecutive pairs [tex] \frac{1}{3k+2}, \frac{-1}{3(k+1) } [/tex] will "telescope", leaving a small positive residue term each time. Then [tex] S > \sum_{k=1}^{\infty} \frac{1}{3k+1} > \sum_{k=1}^{\infty} \frac{1}{3(k+1)} = \frac{1}{3} \sum_{k=2}^{\infty} \frac{1}{k} [/tex] .
 

1. What is the convergence of a series?

The convergence of a series refers to the behavior of the terms in the series as more and more of them are added together. A series is said to converge if the sum of its terms approaches a finite value as the number of terms increases.

2. How do you determine if a series converges or diverges?

There are several tests that can be used to determine the convergence or divergence of a series, including the ratio test, the root test, and the comparison test. These tests involve analyzing the behavior of the terms in the series and their relationship to other known convergent or divergent series.

3. What is the difference between absolute and conditional convergence?

Absolute convergence refers to a series that converges regardless of the order in which the terms are added, while conditional convergence refers to a series that only converges when the terms are added in a specific order. A series that is absolutely convergent is also conditionally convergent, but the reverse is not always true.

4. Can a divergent series be manipulated to make it converge?

No, a divergent series cannot be manipulated to make it converge. The convergence of a series is determined by the behavior of its terms, and no amount of manipulation can change this behavior.

5. Why is understanding convergence of series important?

Understanding convergence of series is important in many fields of science, including physics, engineering, and mathematics. It allows us to determine the behavior of infinite sums and make accurate predictions about the behavior of systems that involve continuously changing values. Additionally, many important mathematical concepts, such as integration and differentiation, rely on understanding the convergence of series.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
691
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
687
  • Calculus and Beyond Homework Help
Replies
1
Views
200
  • Calculus and Beyond Homework Help
Replies
7
Views
938
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
29
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
770
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
Back
Top