Showing that ratio $(4n^2 + 4n + 1)^n$ to $(4n^2 + 4n)^n$ is approximately $4n + 4$ to $4n + 3$

Question

After messing around with some computer simulations, I've come to the observation that the ratio of $(4n^2 + 4n + 1)^n$ to $(4n^2 + 4n)^n$ is approximately $4n + 4$ to $4n + 3$. But I'm not sure how to show this mathematically. I've tried binomial expanding both, but there's just too many terms that don't seem to simplify, so maybe there is a clever way to see this without having to do too much algebra? Any hints would be well-appreciated.

Answer 1

Let $$a_n=\left(\frac{4 n^2+4 n+1}{4 n^2+4 n}\right)^n\implies\log(a_n)=n \log\left(1+\frac{1}{4 n^2+4 n}\right)$$ Using the classical expansion of $\log(1+x)$, replace $x$ by $\frac{1}{4 n^2+4 n}$ and continue with Taylor series to make $$\log(a_n)=n \Bigg[\frac{1}{4 n^2}-\frac{1}{4 n^3}+\frac{7}{32 n^4}+O\left(\frac{1}{n^5}\right) \Bigg]$$ $$\log(a_n)=\frac{1}{4 n}-\frac{1}{4 n^2}+\frac{7}{32 n^3}+O\left(\frac{1}{n^4}\right)$$ Now $$a_n=e^{\log(a_n)}=1+\frac{1}{4 n}-\frac{7}{32 n^2}+O\left(\frac{1}{n^3}\right)$$ While $$\frac{4 n+4}{4 n+3}=1+\frac{1}{4 n}-\frac{3}{16 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\left(\frac{4 n^2+4 n+1}{4 n^2+4 n}\right)^n-\frac{4 n+4}{4 n+3}=-\frac{1}{32n^2}+O\left(\frac{1}{n^3}\right)$$

If you make $n=10$ the exact value is $$\frac{278218429446951548637196401}{271973609384181760000000000}- \frac{44}{43}=$$ $$-\frac{3446346685080848600554757}{11694865203519815680000000000}=-0.000294689\cdots$$ while $$-\frac{1}{3200}=-0.0003125$$

Pretty good approximation, for sure.

Edit

We can improve writing $$\left(\frac{4 n^2+4 n+1}{4 n^2+4 n}\right)^n-\frac{4n+\alpha}{4n+\beta}=\frac{-\alpha +\beta +1}{4 n}+\frac{2 \beta (\alpha -\beta )-7}{32 n^2}+\frac{6 \beta ^2 (\beta -\alpha )+61}{384 n^3}+O\left(\frac{1}{n^4}\right)$$ which give $\alpha=\frac 92$ and $\beta=\frac 72$ $$\left(\frac{4 n^2+4 n+1}{4 n^2+4 n}\right)^n-\frac{8n+9}{8 n+7}=-\frac{25}{768 n^3}+O\left(\frac{1}{n^4}\right)$$ which reduces the error by a factor of $\frac{24 }{25}n$.

Making as before $n=10$ $$\frac{278218429446951548637196401}{271973609384181760000000000}- \frac{89}{87}=-0.0000273807$$ while $$-\frac{25}{768000}=-\frac{1}{30720}=-0.0000325521$$