I'm trying to prove this equality (unsuccessfully): $$\sum_{n=0}^\infty \frac{i^nt^n}{n!}=\sum_{n=0}^\infty \frac{(-1)^nt^{2n}}{(2n)!}+i\sum_{n=0}^{\infty}\frac{(-1)^nt^{2n+1}}{(2n+1)!}\ , $$ which I know is true because: the left hand side is $e^{it}$ while the right hand side is $\cos t+i\sin t$.
I have tried to manipulate the RHS of several ways, but I can't see the trick! Thanks in advance.
For every integer $j \geq 0$, $i^{2j} = (-1)^j$ and $i^{2j+1} = (-1)^j i$
Now, separate out the terms with even $n$ and odd $n$; i.e., when $n = 2j$ and when $n = 2j + 1$.
$$\sum_{n=0}^\infty \frac{i^nt^n}{n!} = \sum_{n=0, n \text{ even}}^\infty \frac{i^nt^n}{n!} + \sum_{n=0, n \text{ odd}}^\infty \frac{i^nt^n}{n!}$$
$$ = \sum_{j=0}^\infty \frac{i^{2j}t^{2j}}{(2j)!} + \sum_{j=0}^\infty \frac{i^{2j+1}t^{2j+1}}{(2j+1)!}$$
$$ = \sum_{j=0}^\infty (-1)^j\frac{t^{2j}}{(2j)!} + \sum_{j=0}^\infty i(-1)^j\frac{t^{2j+1}}{(2j+1)!}$$
$$ = \sum_{j=0}^\infty (-1)^j\frac{t^{2j}}{(2j)!} + i\sum_{j=0}^\infty (-1)^j\frac{t^{2j+1}}{(2j+1)!}$$
All that is left to do is to change the summation index from $j$ back to $n$
Think of the first few terms of your series:
$1 + it - \frac{t^2}{2!} - i\frac{t^3}{3!} + \frac{t^4}{4!} + i\frac{t^5}{5!} - \frac{t^6}{6!} - i\frac{t^7}{7!} + \cdots$
We've evaluated the powers of $i$ appearing in each term. If you now separate the real and imaginary terms into separate sums, you'll have all the even terms in the real sum, and all the odd terms in the imaginary one. That is precisely what the RHS represents.
