Convergent or Divegent Series?

In summary, the conversation discusses the convergence or divergence of the series \sum{(ln(k))/(\sqrt{k+2})} as k approaches infinity. The speaker initially believes the series to converge by using L'Hopital's rule and their calculator, but is reminded by others that the definition of a series involves taking the limit of partial sums. The conversation then shifts to comparing the series to \frac{1}{\sqrt{k+2}} to determine its convergence, with the speaker receiving a helpful hint.
  • #1
iatnogpitw
9
0

Homework Statement


[tex]\sum{(ln(k))/(\sqrt{k+2})}[/tex], with k starting at 1 and going to [tex]\infty[/tex]


Homework Equations


Does this series converge or diverge? Be sure to explain what tests were used and why they are applicable.


The Attempt at a Solution


Okay, my TA got that this diverges, but I got that it converges by simply taking the limit as k goes to [tex]\infty[/tex] and applying L'Hopital's rule. I also plugged the function into my calculator and it seems to converge at y=0, which is corroborates what I got with L'Hopitals rule. What did you guys get? Any help is greatly appreciated.
 
Physics news on Phys.org
  • #2
It looks like you proved the individual terms go to 0, which isn't what you're trying to do. The definition of a series is you take the limit of the partial sums. And where does y come into the series?

Try comparing the series to [tex] \frac{1}{\sqrt{k+2}}[/tex] whose convergence/divergence is easier to find
 
  • #3
Right, I forgot about the partial sums. Thanks, that helped a lot. But isn't [tex]1/(\sqrt{k+2})[/tex] smaller than [tex](ln(k))/(\sqrt{k+2})[/tex]?
 
Last edited:
  • #4
iatnogpitw said:
Right, I forgot about the partial sums. Thanks, that helped a lot. But isn't [tex]1/(\sqrt{k+2})[/tex] smaller than [tex](ln(k))/(\sqrt{k+2})[/tex]?
Yeah, it is. You've been handed a clue for free. If you can say something about what [tex]\sum 1/(\sqrt{k+2})[/tex] does, then maybe you will know something about the series you're really interested in.
 

1. What is the definition of a convergent series?

A convergent series is a sequence of numbers that approaches a finite limit as the number of terms increases. In other words, as more terms are added to the series, the sum of those terms approaches a specific value.

2. How can you determine if a series is convergent or divergent?

One way to determine the convergence or divergence of a series is by using the ratio test. This test involves taking the limit of the absolute value of the ratio between consecutive terms in the series. If this limit is less than 1, the series is convergent. If the limit is greater than 1, the series is divergent. If the limit is equal to 1, the test is inconclusive and another method must be used.

3. What is the difference between a convergent and divergent series?

A convergent series has a finite limit, meaning that the sum of all the terms in the series approaches a specific value. In contrast, a divergent series does not have a finite limit and the sum of its terms either approaches infinity or oscillates between positive and negative infinity.

4. Can a series be both convergent and divergent?

No, a series cannot be both convergent and divergent. By definition, a series can only have one of these two properties. However, a series can be conditionally convergent, meaning that it is convergent when certain conditions are met, but divergent when those conditions are not met.

5. How can you use the sum of a convergent series to find the sum of a divergent series?

The sum of a divergent series cannot be found using the same methods as a convergent series. However, some divergent series can be manipulated or transformed into a convergent series by rearranging the terms or using other mathematical techniques. In these cases, the sum of the new convergent series can be used as an approximation for the sum of the original divergent series.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
165
  • Calculus and Beyond Homework Help
Replies
2
Views
718
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
699
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
398
  • Calculus and Beyond Homework Help
Replies
29
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
473
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
Back
Top