convergence and divergence

Philip K.

New member
Joined
Apr 28, 2020
Messages
10
I have the following problem: Proof if it is convergence or divergence.
I am trying to do it using the so-called " ratio test "

\(\displaystyle
\sum ^{\infty }_{n=1}\dfrac {n^{n}\cdot n!}{\left( 2n\right) !}
\)

\(\displaystyle
a_n=\dfrac {n^{n}\cdot n!}{\left( 2n\right) !}
\)
\(\displaystyle
a_{n+1} = \dfrac {\left( n+1\right) ^{\left( n+1\right) }.\left( n+1\right) !}{\left( 2\left( n+1\right) \right) !}
\)

\(\displaystyle
\lim _{n\rightarrow \infty }\left[ \dfrac {\left( n+1\right) ^{\left( n+1\right) }\left( n+1\right) !}{\left( 2\left( n+1\right) \right) !}\cdot \dfrac {2n!}{n^{n}\cdot n!}\right]
\)

\(\displaystyle
\lim _{n\rightarrow \infty }\left[ \dfrac {\left( n+1\right) ^{n}\left( n+1\right) \cdot \left( n+1\right) n!}{2.n!\left( n+1\right) }\cdot \dfrac {2n!}{n^{n}\cdot n!}\right]
\)

\(\displaystyle
\lim _{n\rightarrow\infty }\left[ \dfrac {\left( n+1\right) ^{n}\cdot \left( n+1\right) }{n^{n}}\right]
\)

Here I am kind of stuck with the calculation of the limit.

One more similar problem where I am stuck in the same place is :
\(\displaystyle
\sum ^{\infty }_{n=1}\dfrac {n^{n}}{2^{n}\cdot n!}
\)

Can someone explain how should i proceed?
 
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Jomo

Elite Member
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Dec 30, 2014
Messages
8,376
You did not divide the factorials correctly.

(2(n+1))! does not become 2*(n+1)!

Also in the limit you did end up with you should know that (n+1)^2/n^n goes to 1!
 

Philip K.

New member
Joined
Apr 28, 2020
Messages
10
You did not divide the factorials correctly.

(2(n+1))! does not become 2*(n+1)!

Also in the limit you did end up with you should know that (n+1)^2/n^n goes to 1!
I have written 2*n!(n+1) though, not 2*(n+1)! .
 

Jomo

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Dec 30, 2014
Messages
8,376

Singleton

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Jun 20, 2020
Messages
49
Before you use a test for convergence, you should always check that the terms at least go to zero. Otherwise there is no chance. When n=1000, is
a_n really small? You want to have a knowledge of the order of magnitude of these functions. Then you will have a good idea what to expect and how to proceed in proving what you expect.
 

pka

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Joined
Jan 29, 2005
Messages
10,453
I have the following problem: Proof if it is convergence or divergence.
\(\displaystyle \sum ^{\infty }_{n=1}\dfrac {n^{n}\cdot n!}{\left( 2n\right) !}\)
Here I am kind of stuck with the calculation of the limit.
Can someone explain how should i proceed?
Here are some hints. None of the elementary tests will work.
However, there is a basic comparison test that will answer the question.
If you find it, please post it.
 

Philip K.

New member
Joined
Apr 28, 2020
Messages
10
Here are some hints. None of the elementary tests will work.
However, there is a basic comparison test that will answer the question.
If you find it, please post it.
Still cannot figure it out though...
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
8,376
Use pka's hint!

Can you simplify nn*n!/(2n)! ?
 

lookagain

Elite Member
Joined
Aug 22, 2010
Messages
2,639
I have the following problem: Proof if it is convergence or divergence.
I am trying to do it using the so-called " ratio test "

Can someone explain how should i proceed?
You can continue with the ratio test as an option (and use post #4).

I want to save on typing with my tablet. I'll use N for the numerator
and D for the denominator :

N: \(\displaystyle (n + 1)^n(n + 1)*(n + 1)n!*(2n)!\)
D: \(\displaystyle (2n + 2)!*n^n*n!\)


D: \(\displaystyle (2n + 2)(2n + 1)(2n)!*n^n*n!\)
D: \(\displaystyle 2(n + 1)(2n + 1)(2n)!*n^n*n!\)

Reduce previous numerator with the latest denominator:

N: \(\displaystyle (n + 1)^n*(n + 1)\)
D: \(\displaystyle 2n^n*(2n + 1)\)

D: \(\displaystyle 2*2n^n*(n + 1/2)\)

Put the N and the D together, but as a product of fractions:

\(\displaystyle \dfrac{1}{4}*\dfrac{(n + 1)^n}{n^n}*\dfrac{n + 1}{n + 1/2} \ \ \ \ \) . . . . . . .**

Rewrite the middle as:

\(\displaystyle \bigg(1 + \dfrac{1}{n} \bigg)^n\)

Now, take the limit as n approaches infinity of the absolute
value of this product **.

You need to be able to do this part now and the steps
prior to this written out more fully than I have shown.



If this limit is less than 1, then you know the series is
convergent, for example.
 
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