Determining whether the series is convergent or divergent

[umzung]

Homework Statement

Determine if the series is convergent.

Homework Equations

∞
∑ (((2n^2 + 1)^2)*4^n)/(2(n!))
n=1[/B]

The Attempt at a Solution

I'n using the Ratio Test and have got as far as (4*(2(n+1)^2+1)^2)/((n+1)((2n^2+1)^2)). I know this series converges but I need to find the limit to be < 1 to show this. Is there way to now divide each term by the dominant term n^2, or do I need to multiply the whole thing out and divide by the new dominant term? I've tried that and have found the limit to be 0/4 = 0.

 

[Orodruin]

Just note that the numerator goes as $8n^4 + \mathscr O(n^3)$ and that the denominator goes as $4n^5 + \mathscr O(n^4)$ as $n\to \infty$. This will be sufficient.

 

[vela]

Homework Statement

Determine if the series is convergent.

Homework Equations

∞
∑ (((2n^2 + 1)^2)*4^n)/(2(n!))
n=1[/B]

The Attempt at a Solution

I'n using the Ratio Test and have got as far as (4*(2(n+1)^2+1)^2)/((n+1)((2n^2+1)^2)). I know this series converges but I need to find the limit to be < 1 to show this. Is there way to now divide each term by the dominant term n^2, or do I need to multiply the whole thing out and divide by the new dominant term? I've tried that and have found the limit to be 0/4 = 0.

If you want to show the algebra, you can do something like this:
$$\frac{4}{n+1}\frac{[2(n+1)^2+1]^2}{(2n^2+1)^2}
=\frac{4}{n+1}\frac{\left[(n+1)^2\left(2+\frac{1}{(n+1)^2}\right)\right]^2}{\left[n^2\left(2+\frac{1}{n^2}\right)\right]^2}$$ so you don't have to multiply everything out.