Divergence/Convergency of a series

  • Thread starter Heaviplace
  • Start date
  • Tags
    Series
In summary, the homework statement states that the student needs to test a series for convergentity/divergence. They go on to say that the series (3/4)^n - n/(n+10000) does not converge by the root test, but does converge by the limit test. However, the student is still having trouble due to the sum of a convergent series and a divergent series always diverging.
  • #1
Heaviplace
4
0

Homework Statement



I have to test this series for convergency/divergency

Ʃ1000(3/4)^n - n/(n+10000)

n=1 to ∞

The Attempt at a Solution



At first I thought of analyzing both parts separately, (3/4)^n is a geometric series with r < 1 and so it converges, and applying the root test to n/(n+10000). Then I'd have the sum of two convergent series that ofc, also converges.

Went to test my hypothesis on WolframAlpha, and it actually told me the series diverges by the limit test, which I verified to be true.

So my question is, why was the method I used, wrong?

Thanks in advance!
 
Physics news on Phys.org
  • #2
n/(n+10000) doesn't converge by the root test. It doesn't converge at all. It doesn't even approach zero!
 
  • #3
Indeed, you are right!

But that still leaves me the problem, but this time with the sum of a convergent series and divergent series =/
 
  • #4
Heaviplace said:
Indeed, you are right!

But that still leaves me the problem, but this time with the sum of a convergent series and divergent series =/

The sum of a convergent series and a divergent series always diverges. Think of the partial sums.
 
  • #5
Ok, see if I got it right, since one of the series does not converge, I can't bring up the property of the sum of two convergent series.

And even though (3/4)^n is convergent, the limit test shows that the sum of both series (<>0) is not.
 
  • #6
Thanks really man, you really cleared things up for me.

Just a last question a bit off topic, where's the edit button? =P
 
  • #7
Heaviplace said:
Ok, see if I got it right, since one of the series does not converge, I can't bring up the property of the sum of two convergent series.

And even though (3/4)^n is convergent, the limit test shows that the sum of both series (<>0) is not.

Sure, you can think of it that way. (3/4)^n approaches zero as n->infinity. n/(n+10000) doesn't. So the sum doesn't approach zero. Doesn't converge by the limit test.
 
  • #8
Heaviplace said:
Thanks really man, you really cleared things up for me.

Just a last question a bit off topic, where's the edit button? =P

The 'Edit' button has been coming and going lately. No idea why.
 

1. What is the definition of divergence in a series?

Divergence in a series means that the terms in the series are increasing or decreasing without bound, resulting in an infinite sum.

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

A series can be determined to be convergent or divergent by using various tests such as the ratio test, integral test, or comparison test.

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

Absolute convergence means that the series converges regardless of the order of the terms, while conditional convergence means that the series only converges when the terms are arranged in a specific order.

4. Can a series be both convergent and divergent?

No, a series can only be either convergent or divergent. If a series is convergent, it cannot also be divergent.

5. What happens if a series is neither convergent nor divergent?

If a series is neither convergent nor divergent, it is considered to be oscillating or alternating. This means that the series does not approach a specific value and instead fluctuates between positive and negative values.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
189
  • Calculus and Beyond Homework Help
Replies
1
Views
259
  • Calculus and Beyond Homework Help
Replies
2
Views
711
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
958
  • Calculus and Beyond Homework Help
Replies
7
Views
709
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
418
  • Calculus and Beyond Homework Help
Replies
5
Views
991
Back
Top