Natural log : Has this proof jumped a step? This proof in Nevanlinna and Paatero's Introduction to Complex Analysis (1969) is meant to show that if $$c_n=\left(1+\frac{z}{n}\right)^n$$ $(n=1,2,\dots,$  and $z = x + iy)$ then $$e^x = \liminf_ {n}|c_n|$$ ($|.|$ denotes the absolute value (corrected from "the real part").
How do they get from Step $2$ to Step $3$? It looks as if the log function has just been ignored, but they're too reputable for that!
Expanding $$|c_n|=\left\lvert1+\frac{x}{n}+i\frac{y}{n}\right\rvert^n
=\bigg(\left(1+\frac{x}{n}\right)^2+\left(\frac{y}{n}\right)^2\bigg)^\frac{n}{2}$$
so $$\ln|c_n|=\frac{n}{2}\ln\bigg(1+2\frac{x}{n}+\frac{x^2+y^2}{n^2}\bigg)\tag 1$$
$$=\frac{n}{2}\bigg(\frac{2x}{n}+\left[\frac{1}{n^2}\right]\bigg) = x + \left[\frac{1}{n}\right]\tag 2$$
($[.]$ meaning roughly "of the order of").
So $$\liminf_{n}\ln|c_n|=x\tag 3$$  and $$|c_n|\rightarrow e^x$$ QED.

This may be the least troublesome derivation.
For $t > 0,$
$$
\log(1 + t) = \int_1^{1 + t}\frac{du}u < \int_1^{1 + t}du = t,
$$
and
$$
\log(1 + t) = \int_1^{1 + t}\frac{du}u >
\int_1^{1 + t}\frac{du}{1 + t} = \frac{t}{1 + t} =
t - \frac{t^2}{1 + t} > t - t^2,
$$
so
$$
0 < t - \log(1 + t) < t^2.
$$
We don't need the detailed inequality, only the consequential
estimate:
$$
\log(1 + t) = t + O(t^2).
$$
This gives us:
\begin{align*}
\log\left(1 + \frac{2x}n + \frac{x^2 + y^2}{n^2}\right) & =
\log\left(1 + \frac{2x}n + O\left(\frac1{n^2}\right)\right) \\
& = \left(\frac{2x}n + O\left(\frac1{n^2}\right)\right) +
O\left(\left(\frac{2x}n + O\left(\frac1{n^2}\right)\right)^2\right)
\\ & =
\frac{2x}n + O\left(\frac1{n^2}\right) + O\left(\frac1{n^2}\right)
\\ & = \frac{2x}n + O\left(\frac1{n^2}\right),
\end{align*}
whence:
$$
\frac{n}2\log\left(1 + \frac{2x}n + \frac{x^2 + y^2}{n^2}\right) =
x + O\left(\frac1n\right).
$$
