# Thread: Detrmine whthere the sequence converges

1. ## Detrmine whthere the sequence converges

An = 2((n+1)!) + 4^n / 2(n!)n^2

1) determine whether it converges and if it does, find the limit of the sequence {An}

I realise that i need to find the dominant term of the sequnece and to divide by this and so on, but i have not seen an example where (n+1)! is an element of the sequence and this has thrown me. Any help would be highly appreciated!! Limits is not my forte!

2. ## Re: Detrmine whthere the sequence converges

Originally Posted by raggie29
An = 2((n+1)!) + 4^n / 2(n!)n^2

1) determine whether it converges and if it does, find the limit of the sequence {An}

Which is it
$\frac{2(n+1)!+4^n}{2(n!)n^2}\text{ or }2(n+1)!+\frac{4^n}{2(n!)n^2}$

3. ## Re: Detrmine whthere the sequence converges

The first one. Thanks.

4. ## Re: Detrmine whthere the sequence converges

Originally Posted by raggie29
The first one. Thanks.

Why don't you bother to use grouping symbols?
The way you posted the expression, it actually reads as the second one.

$\frac{2(n+1)!+4^n}{2(n!)n^2}=\frac{2(n+1)+\tfrac{4 ^n}{n!}}{2n^2}$

What can you say about $\left(\frac{4^n}{n!}\right)\to~?$

5. ## Re: Detrmine whthere the sequence converges

It is of form of a basic null sequence and hence = 0?

6. ## Re: Detrmine whthere the sequence converges

Originally Posted by raggie29
It is of form of a basic null sequence and hence = 0?
So what is the overall answer?

3/2n ??

8. ## Re: Detrmine whthere the sequence converges

Or is it, = 0? since 1/n and 1/n^2 are both basic null sequences?