D'Alembert criterion

Question

can someone explain to me the middle steps between (1), (2), (3)? This exercise is used as an example of the convergence of series, and I understand the criterion. I don't follow the middle steps.

Thank you in advance.

Answer 1

The fraction above (1) can be rewritten as $$ \left|\frac{(n+1)^3+\sqrt{n+1}}{(2+4i)^{n+1}}\frac{(2+4i)^n}{n^3+\sqrt{n}}\right|= \frac{(n+1)^3+\sqrt{n+1}}{n^3+\sqrt{n}}\frac{1}{|2+4i|} $$ In your image the numerator and denominator are wrongly swapped. The absolute value around the first fraction in the right-hand side is redundant.

It's easy to prove that $$ \lim_{n\to\infty}\frac{(n+1)^3+\sqrt{n+1}}{n^3+\sqrt{n}}=1 $$ so we have $$ \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}=\frac{1}{|2+4i|}=\frac{1}{\sqrt{20}} $$

Answer 2

$$\lim_{n\to \infty}{\left|{(n+1)^3+\sqrt{n+1}\over (2+4i)^{n+1}}\right|\over \left|{n^3+\sqrt{n}\over (2+4i)^{n}}\right|}{=\lim_{n\to \infty}{\left|{(n+1)^3+\sqrt{n+1}}\right|\over \left|{n^3+\sqrt{n}}\right|}\cdot \left|{{1\over (2+4i)^{n+1}}\over {1\over (2+4i)^{n}}}\right|\\=\lim_{n\to \infty}{\left|{(n+1)^3+\sqrt{n+1}}\right|\over \left|{n^3+\sqrt{n}}\right|}\cdot \left|{{(2+4i)^{n}\over (2+4i)^{n+1}}}\right|\\=\lim_{n\to \infty}{\left|{(n+1)^3+\sqrt{n+1}}\right|\over \left|{n^3+\sqrt{n}}\right|}\cdot {1\over |2+4i|}\\=1\cdot {1\over \sqrt{20}}\\={1\over \sqrt{20}}}$$