Convergence/Divergence of series

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In summary, the conversation discusses using the ratio test to prove convergence or divergence of a given series. However, upon using the ratio test, it is found that the series diverges, despite evidence of it converging online. The conversation then delves into possible mistakes or errors in the process, including the use of Stirling's approximation and factoring out terms in the series. Ultimately, it is concluded that the series is more likely to diverge due to the high values of each term.
  • #1
Thomas_
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Hello,

I have to prove conv/div. for the following series:

[tex]\sum\frac{(2n)!}{n^n}[/tex]

I use the "ratio-test" and get the following:

[tex]\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}} = \lim_{n\to\infty} \frac{(2n+2)!}{(2n)!} \frac{n^n}{(n+1)^{n+1}} = \lim_{n\to\infty} \frac{(2n+2)(2n+1)}{(n+1)} (\frac{n}{1+n})^n = \infty \frac{1}{e} = \infty[/tex]

This means the series diverges, however, the series should converge (I could find the finite sum online).

Where is my mistake?

Thank you!
 
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  • #2
The last half terms can be written as {(n+l)/n}{(n+2)/n}{(n+3)/n}...{2n/n}, and the first n terms are just n!.
 
  • #3
robert Ihnot said:
The last half terms can be written as {(n+l)/n}{(n+2)/n}{(n+3)/n}...{2n/n}, and the first n terms are just n!.
Sorry, I do not quite understand what you mean or how this helps me. Could you elaborate on that?

Also, I am interested in why the test I am using does not work out like it should or if I made an algebra mistake somewhere along the way.
 
  • #4
Using stirlings approximation to replace the factorial, I get the series diverges. Where did you find online its sum?
 
  • #5
Thomas_ said:
Sorry, I do not quite understand what you mean or how this helps me. Could you elaborate on that?

Also, I am interested in why the test I am using does not work out like it should or if I made an algebra mistake somewhere along the way.

What he's saying is that if you split it up, you get 1/n*1/n*1/n...*(2n)(2n-1)(2n-2)...(n+1)*n!

So you put one n under each 2n-k and get

2n/n*(2n-1)/n*(2n-2)/n...*(n+1)/n*n!

As each (2n-k)/n>1, and n!>1, each term in the series is >1. So there's very little reason why it would converge
 

FAQ: Convergence/Divergence of series

What is the definition of convergence and divergence of a series?

The convergence of a series refers to the behavior of the sum of its terms as the number of terms increases towards infinity. A series is said to converge if the sum of its terms approaches a finite limit. On the other hand, a series is said to diverge if the sum of its terms does not approach a finite limit and instead tends towards infinity.

How do you determine if a series is convergent or divergent?

There are several tests that can be used to determine if a series is convergent or divergent. Some of the most commonly used tests include the comparison test, ratio test, root test, and the integral test. These tests involve comparing the given series to a known convergent or divergent series, or evaluating the limit of a certain function related to the series. If the limit is finite, the series is convergent. If the limit is infinite, the series is divergent.

What is the significance of the convergence or divergence of a series?

The convergence or divergence of a series is important in mathematics as it determines whether the sum of the terms of the series will approach a finite value or not. This has implications in various areas of mathematics, such as calculus, where the convergence or divergence of a series can determine the behavior of a function. In other fields, such as physics and engineering, the convergence or divergence of a series can help in modeling and predicting real-world phenomena.

Can a series be both convergent and divergent?

No, a series can only be either convergent or divergent. If a series is convergent, it means that the sum of its terms approaches a finite value. If a series is divergent, it means that the sum of its terms does not approach a finite value and instead tends towards infinity. These two behaviors are mutually exclusive, and a series cannot exhibit both at the same time.

How does the rate of convergence or divergence affect the behavior of a series?

The rate of convergence or divergence of a series can determine how quickly or slowly the sum of its terms approaches a finite value or tends towards infinity. In general, a series with a faster rate of convergence will approach its limiting value more quickly than a series with a slower rate of convergence. Similarly, a series with a slower rate of divergence will tend towards infinity at a slower pace than a series with a faster rate of divergence.

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