TITLE: Deriving Taylor series without applying Taylor's theorem.
QUESTION [6 upvotes]: First, a neat little 'proof' of the Taylor series of $e^x$.
Start by proving with L'Hospital's rule or similar that
$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
and then binomial expand into
$$e^x=\lim_{n\to\infty}1+x+\frac{n-1}n\frac{x^2}2+\dots+\frac{(n-1)(n-2)(n-3)\dots(n-k+1))}{n^{k-1}}\frac{x^k}{k!}+\dots$$
Evaluating the limit, we are left with
$$e^x=1+x+\frac{x^2}2+\dots+\frac{x^k}{k!}+\dots$$
which is our well known Taylor series of $e^x$.
As dxiv mentions, we can exploit the geometric series:
$$\frac1{1-x}=1+x+x^2+\dots$$
$$\ln(1+x)=x-\frac12x^2+\frac13x^3-\dots$$
$$\arctan(x)=x-\frac13x^3+\frac15x^5-\dots$$
which are found by integrating the geometric series and variants of it.
I was wondering if other known Taylor series can be created without applying Taylor's theorem.  Specifically, can we derive the expansions of $\sin$ or $\cos$?

REPLY [4 votes]: [A].From "101 Great Problems In Elementary Mathematics" by H. Dorrie : For $x\geq 0$ we have 
(1).$\;\sin x \leq x\implies 1-\cos x=\int_0^x\sin y \;dy\leq \int_0^xy\;dy=x^2/2\implies \cos x\geq 1-x^2/2!.$
(2). From (1), $\;\sin x=\int_0^x\cos y\;dy\geq \int_0^y(1-y^2/2!)dy=x-x^3/3!.$
(3).From (2), $1-\cos x=\int_0^x\sin y dy\geq \int_0^x(y-y^3/3!)dy=y^2-y^4/4!\implies \cos x\leq 1-x^2/2!+x^4/4!.$ 
Et cetera.
(4). So in general for $n>0$ we  have $$\sum_{j=1}^{2n}(-1)^{j-1}x^{2j-1}/(2j-1)!\leq \sin x \leq\sum_{j=1}^{2n-1}(-1)^{j-1}x^{2j-1}/(2j-1)!$$ and a similar set of inequalities for $\cos x$.
This gives us the power series for $\sin x$ and $\cos x$ for $x\geq 0.$ Since $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x,$ this gives us their power series for all real $x.$
[B]. For $x\geq 0:$
(1). $e^{-x}\leq 1\implies 1-e^{-x}=\int_0^xe^{-y}dy\leq \int_0^x1dy=x\implies e^{-x}\geq 1-x.$
(2).From (1), $$1-e^{-x}=\int_0^xe^{-y}dy\geq \int_0^x(1-y)dy=y-y^2/2!$$ $$\implies e^{-x}\leq 1-x+x^2/2!.$$
(3). From (2), $$1-e^{-x}=\int_0^xe^{-y}dy\leq \int_0^x(1-y+y^2/2!)dy=x-x^2/2!+x^3/3!$$ $$\implies e^{-x}\geq 1-x+x^2/2!-x^3/3!.$$ Et cetera. 
This gives us the power series for $e^{-x}$ for $x\geq 0.$ And it is an elementary exercise to show that $\lim_{n\to \infty}(\sum_{j=0}^nx^{-j}/j!)(\sum_{j=0}^nx^j/j!)=1,$ so we obtain the series for $e^x$ for all real $x.$