Can somebody explain to me why these terms are equal? I just read a proof on ProofWiki that proves Euler's formula, but I can't seem to understand what is done in this following step:
$$\sum\limits_{n=0}^\infty\left(\frac{(i\theta)^{2n}}{(2n)!}+\frac{(i\theta)^{2n+1}}{(2n+1)!}\right) =\sum\limits_{n=0}^\infty{\frac{(i\theta)^n}{n!}}$$
Could anyone help me understand this equality step by step?
 A: Sure. You understand, I assume, that the first term is the $\cos$ function, and the second term is the $\sin$ function. Try plugging in values of $n$, and see what emerges.
$$n=0: \frac{(i\theta)^{2\cdot0}}{(2\cdot0)!}=\frac{(i\theta)^{0}}{0!}$$
$$n=0: \frac{(i\theta)^{2\cdot0+1}}{(2\cdot0+1)!}=\frac{(i\theta)^{1}}{1!}$$
$$n=1: \frac{(i\theta)^{2\cdot1}}{(2\cdot1)!}=\frac{(i\theta)^{2}}{2!}$$
$$n=1: \frac{(i\theta)^{2\cdot1+1}}{(2\cdot1+1)!}=\frac{(i\theta)^{3}}{3!}$$
Notice the pattern?
$$\frac{(i\theta)^{n}}{n!}$$
The first term ($\cos$) provides the even terms in the second sequence, and the second term ($\sin$) provides the odd terms in the second sequence. Since we are summing over an infinite sequence of integers, the upper bound doesn't have to change.
A: $$
\begin{array}{c|c|c|c}
n & & 2n & 2n+1 \\
\hline
0 & & 0 & 1 \\
1 & & 2 & 3 \\
2 & & 4 & 5 \\
3 & & 6 & 7 \\
4 & & 7 & 8 \\
5 & & 8 & 9 \\
6 & & 12 & 13 \\
7 & & 14 & 15 \\
\vdots & & \vdots & \vdots
\end{array}
$$
As $n$ goes through the list $0,1,2,3,4,\ldots$, the two later columns together also go through the list $0,1,2,3,4,\ldots$, each column going through half of it.  Therefore
$$
\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty (a_{2n} + a_{2n+1}).
$$
A: If you write out the first few terms on each side, you will see that the left side groups pairs of terms from the right side.
The idea is:
$$(a_0 + a_1) + (a_2 + a_3) + \cdots = a_0+a_1+a_2+a_3 + \cdots$$
A: Just expand
$$\sum\limits_{n=0}^\infty\left(\frac{(i\theta)^{2n}}{(2n)!}+\frac{(i\theta)^{2n+1}}{(2n+1)!}\right) =\left(\frac{1}{0!}+\frac{i\theta}{1!}\right)+\left(\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}\right)+\left(\frac{(i\theta)^4}{4!}+\frac{(i\theta)^5}{5!}\right)+\dotsb .$$ Now assuming all the nice convergence properties we can rewrite this (without the parentheses) as 
$$\text{RHS} =\sum\limits_{n=0}^\infty{\frac{(i\theta)^n}{n!}}.$$
