# convergence and divergence

#### Philip K.

##### New member
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
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
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)! .

#### Subhotosh Khan

##### Super Moderator
Staff member
I have written 2*n!(n+1) though, not 2*(n+1)! .
That is not correct either.

[2(n+1)]! = (2n+2)! = (2n+2) * (2n+1) * [(2n)!]

#### Jomo

##### Elite Member
I have written 2*n!(n+1) though, not 2*(n+1)! .
Sorry if I made a mistake, but those two are equal anyways.

#### Singleton

##### New member
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

##### Elite Member
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
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
Use pka's hint!

Can you simplify nn*n!/(2n)! ?

#### lookagain

##### Elite Member
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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