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! :(

Re: Detrmine whthere the sequence converges

Quote:

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

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

Re: Detrmine whthere the sequence converges

The first one. Thanks. :)

Re: Detrmine whthere the sequence converges

Quote:

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.

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

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

Re: Detrmine whthere the sequence converges

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

Re: Detrmine whthere the sequence converges

Quote:

Originally Posted by

**raggie29** It is of form of a basic null sequence and hence = 0?

So what is the overall answer?

Re: Detrmine whthere the sequence converges

Re: Detrmine whthere the sequence converges

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