# Thread: Convergent series by Ratio Test

1. ## Convergent series by Ratio Test

I have to show the following series is either convergent or divergent.

$\displaystyle \sum_0^\infty \frac{(n^2 +1)^2.2^{2n-1}}{n!}$

I know this series is convergent by the ratio test, but I am having trouble showing this proof.

So far I have:

$\displaystyle \rho = lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = lim_{n\rightarrow\infty} \frac{((n+1)^2 +1)^2.2^{2(n+1)-1}}{(n+1)!} . \frac{n!}{(n^2 +1)^2.2^{2n-1}}$

I simplified to
$\displaystyle \rho = lim_{n\rightarrow\infty} \frac{((n+1)^2 +1)^2}{(n^2+1)^2}.\frac{2^{2n+1}}{2^{2n-1}}.\frac{n!}{(n+1)n!}}$
$\displaystyle = lim_{n\rightarrow\infty} \frac{(n^2 +2n+2)^2}{(n^2+1)^2}.2^2.\frac{1}{(n+1)}$
$\displaystyle = lim_{n\rightarrow\infty} \frac{4.(n^2 +2n+2)^2}{(n^2+1)^2(n+1)}$

This is where I am stuck.

Could someone help me check if this is right so far and help me with what comes next?

2. ## Re: Convergent series by Ratio Test

It looks ok to me so far. It might be a good idea to expand all your brackets and then divide top and bottom by the highest power of n...

3. ## Re: Convergent series by Ratio Test

Thanks for the response.

So, expanding the brackets:
$\displaystyle \rho=lim_{n\rightarrow\infty} \frac{4n^4 + 16n^3+32n^2+32n+16}{n^5+n^4+2n^3+2n^3+n+1}$

dividing through by $\displaystyle n^5$ gives:

$\displaystyle \rho=\frac{0}{1}=0<1$

therefore series is convergent.

Is this an appropriate way to complete the problem?

4. ## Re: Convergent series by Ratio Test

$\displaystyle = lim_{n\rightarrow\infty} \frac{4.(n^2 +2n+2)^2}{(n^2+1)^2(n+1)}$
If you were to multiply the terms in numerator out you would see that the numerator has highest power of $\displaystyle n^4$ and the denominator has highest power $\displaystyle n^5$