Solve Serie Convergence: Cauchy, d'Alembert & More

In summary, Homework Equations states that the sum of 1+1/n and 1+2/n is 2, and that it is divergent.
  • #1
Jalo
120
0

Homework Statement



Hi. I'm trying to solve a serie:

Ʃ12n+1*(n+1)! / (n+1)n+1

Homework Equations


The Attempt at a Solution



I tried solving it with Cauchy's method, but it failed. I also tried using d'alembert criterion, which game me the answer 2, so it should be divergent.
However in the solutions it says it's convergent. If anyone could check my solution and see if I did anything wrong or if it's a simple error in the solutions, I'd appreciate!
21e5g5g.png
 
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  • #2
You've made a small mistake in the last step. When you have,

[tex]\lim_{n\to \infty} 2 \left ( \frac{1+1/n}{1+2/n} \right ) ^{n+1}[/tex]

You can't directly write it to be equal to 2,because the term in the bracket is of the indeterminate form [itex]1^{\infty}[/itex]
 
  • #3
Infinitum said:
You've made a small mistake in the last step. When you have,

[tex]\lim_{n\to \infty} 2 \left ( \frac{1+1/n}{1+2/n} \right ) ^{n+1}[/tex]

You can't directly write it to be equal to 2,because the term in the bracket is of the indeterminate form [itex]1^{\infty}[/itex]

Oh, I see... I tought I could say it was equal to 1... Any ideas then on how to solve it?
 
  • #4
Jalo said:
Oh, I see... I tought I could say it was equal to 1... Any ideas then on how to solve it?

Um, that is a standard limit of the form,

[tex]\lim_{x\to \infty} (1+\frac{1}{x})^x[/tex]

or, more generally,

[tex]\lim (1+f(x))^{g(x)}[/tex]
As f(x) tends to zero and g(x) tends to infinity.

Did you not learn about these?
 
  • #5
Infinitum said:
Um, that is a standard limit of the form,

[tex]\lim_{x\to \infty} (1+\frac{1}{x})^x[/tex]

or, more generally,

[tex]\lim (1+f(x))^{g(x)}[/tex]
As f(x) tends to zero and g(x) tends to infinity.

Did you not learn about these?

I don't think so. I studied
[tex]\lim_{x\to \infty} (1+\frac{1}{x})^x[/tex]
and I know it tends to e, but I don't know how it helps me in this case.I can't rewrite
[tex]\lim_{n\to \infty} 2 \left ( \frac{1+1/n}{1+2/n} \right ) ^{n+1}[/tex]
as
[tex]\lim_{x\to \infty} (1+\frac{1}{x})^x[/tex]
 
  • #6
Jalo said:
I don't think so. I studied
[tex]\lim_{x\to \infty} (1+\frac{1}{x})^x[/tex]
and I know it tends to e, but I don't know how it helps me in this case.I can't rewrite
[tex]\lim_{n\to \infty} 2 \left ( \frac{1+1/n}{1+2/n} \right ) ^{n+1}[/tex]
as
[tex]\lim_{x\to \infty} (1+\frac{1}{x})^x[/tex]

You should ask your teacher to explain these types of limits, then.

To find

[tex]\lim_{x\to \infty}(1+f(x))^{g(x)}[/tex]

Such that f(x) tends to 0 and g(x) tends to infinity is,

[tex]e^{\lim{x\to \infty}f(x) \cdot g(x)}[/tex]

The limit of [itex](1+1/x)^x[/itex] as [itex]x\to \infty[/itex] is a particular case(sometimes, even used as the definition of e) of this general form.
 
  • #7
Infinitum said:
You should ask your teacher to explain these types of limits, then.

To find

[tex]\lim_{x\to \infty}(1+f(x))^{g(x)}[/tex]

Such that f(x) tends to 0 and g(x) tends to infinity is,

[tex]e^{\lim{x\to \infty}f(x) \cdot g(x)}[/tex]

The limit of [itex](1+1/x)^x[/itex] as [itex]x\to \infty[/itex] is a particular case(sometimes, even used as the definition of e) of this general form.

Thank you very much :)
 
  • #8
Hi Jalo, (Hi Infinitum)
First, you could have made your life easier by starting with a simpler index
(you have n+2 terms over n+1 terms that you could instead put as n+1 terms over n terms)
Then that would lead you to think about the limit of 2(1+1/n)^n
I'm saying think about instead of find out, because as Infinitum told you, this limit can be tricky, and you learned about it or not.
But what is important is the original question, you were not asked to evaluate precisely this limit, you just wanted to check D'Alembert's rule.
Now (1+1/n)^n is hard if you are not properly prepared to deal with it,
but pretend you are not, do you really need to be ?
Whatever this limit is, who knows, it could go to infinity, one thing is certain, it is >=1, that is all you need, since you multiply it by 2, to know that your rule for divergence is satisfied.

Cheers...
 
  • #9
Infinitum said:
Um, that is a standard limit of the form,

[tex]\lim_{x\to \infty} (1+\frac{1}{x})^x[/tex]

or, more generally,

[tex]\lim (1+f(x))^{g(x)}[/tex]
As f(x) tends to zero and g(x) tends to infinity.

Did you not learn about these?

Hello Infinitum,

I have never heard of this. Does it has a particular name? I'm trying to look on the internet, but I don't find anything related to that. Do you have a link that you could share?
 
  • #10
Jalo said:

Homework Statement



Hi. I'm trying to solve a serie:

Ʃ12n+1*(n+1)! / (n+1)n+1

Homework Equations





The Attempt at a Solution



I tried solving it with Cauchy's method, but it failed. I also tried using d'alembert criterion, which game me the answer 2, so it should be divergent.
However in the solutions it says it's convergent. If anyone could check my solution and see if I did anything wrong or if it's a simple error in the solutions, I'd appreciate!
21e5g5g.png

Use Stirling's Formula: [itex] N! \sim \sqrt{2 \pi} N^{N}\sqrt{N} e^{-N}[/itex] for large N. Here, the symbol "~" means that the ratio of the two sides approaches 1 as N approaches infinity.

RGV
 
  • #11
naaa00 said:
Hello Infinitum,

I have never heard of this. Does it has a particular name? I'm trying to look on the internet, but I don't find anything related to that. Do you have a link that you could share?

It is the limit of e only a bit modified.

[tex]\lim_{x\to a} (1+f(x))^{g(x)}[/tex]

Can be re-written as

[tex]\lim_{x\to a} \left (1+f(x))^{\frac{1}{f(x)}} \right)^{g(x)f(x)} [/tex]


I think you should be able to understand it from here
 
  • #12
I see! Really interesting - I actually didn't saw the exponent and the (1/x) inside the parenthesis (for the particular case, e) as to independent functions...
 

FAQ: Solve Serie Convergence: Cauchy, d'Alembert & More

What is the Cauchy criterion for series convergence?

The Cauchy criterion states that if for any positive number ε, there exists a positive integer N such that for all n and m greater than N, the absolute value of the difference between the sum of the first n terms and the sum of the first m terms is less than ε, then the series is convergent.

What is the d'Alembert ratio test for series convergence?

The d'Alembert ratio test states that if the limit of the absolute value of the ratio of consecutive terms in a series is less than 1, then the series is convergent. However, if the limit is greater than 1 or does not exist, then the series is divergent.

How do I use the Cauchy and d'Alembert tests to determine series convergence?

To use the Cauchy criterion, you need to find a positive integer N that satisfies the given condition. To use the d'Alembert ratio test, you need to find the limit of the absolute value of the ratio of consecutive terms. If the limit is less than 1, the series is convergent. If the limit is greater than 1 or does not exist, the series is divergent.

Can a series be both Cauchy and d'Alembert convergent?

Yes, a series can be both Cauchy and d'Alembert convergent. The Cauchy criterion and the d'Alembert ratio test are two different methods of determining series convergence, and if a series satisfies both criteria, then it is considered to be both Cauchy and d'Alembert convergent.

What other tests can be used to determine series convergence?

Other common tests for series convergence include the integral test, comparison test, and the alternating series test. These tests have different conditions and criteria for determining series convergence, and it is important to choose the appropriate test for a given series to determine its convergence.

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