Euler-Maclaurin formula and Stirling's approximation, Spivak Calculus problem 27-19

Question

I am confused by the penultimate part of d (use part (c) to show that... ). The solution is below but I don't see how they get from line 3 to line 4?

Answer 1

Lines 3 and 4 come from separate manipulations of the sequence in line 2, $a_{n} = \frac{n!}{\alpha n^{n+\frac{1}{2}} e^{-n}}$.

To obtain line 3, we square the sequence:

\begin{align*} \lim_{n\rightarrow \infty} \frac{(n!)^2}{\alpha^2 n^{2n+1} e^{-2n}} &= \lim_{n\rightarrow \infty} \left(\frac{n!}{\alpha n^{n+\frac{1}{2}} e^{-n}} \right)^2 \\ &= \left( \lim_{n\rightarrow \infty} a_{n}\right)^2 \\ &= 1 \end{align*}

To obtain line 4, we substitute $2n$ for the index: $a_{2n} = \frac{(2n)!}{\alpha (2n)^{2n + \frac{1}{2}} e^{-2n}}$. So, line 4 just uses the fact that a subsequence of a convergent sequence converges to the same limit.