Obtaining the value of a power series similar to sine I apologies for the vague title and the very specific question. I would like to know what
$$K=4\left[ \frac{1}{1\cdot2!}-\frac{1}{3\cdot4!}+\frac{1}{5\cdot6!}-\cdots \right]$$
evaluates to. This is probably a fairly standard Maths problem, but unfortunately my training extends to identifying sine, cosine, exponentials, and geometric series. What is a good resource to look at?
A related, more general problem is
$$K(a)=4\left[ \frac{a^2}{1\cdot2!}-\frac{a^4}{3\cdot4!}+\frac{a^6}{5\cdot6!}-\cdots \right].$$
Does this one have a closed form? 
Progress (credit to Justpassingby)
This is incorrect, the correct answer is below by kamil09875. However, I wonder if anyone could explain why this is incorrect?
$$K(a)=4a^2\frac{\text{d}}{\text{d}a}\left[\frac{1-\cos(a)}{a}\right]
=4\left[-1+\cos(a)+a\sin{a}\right],$$
which is a closed form!
 A: If you write down the expansion for $\cos a,$ subtract the first term and then divide by $a^2$ you arrive at a power series is the term-wise derivative of (a linear multiple of) this series. So at least the derivative of your series, divided by $a,$ with respect to $a$ has a closed form.
I am not sure if this is useful as a general approach. It is certainly useful to know certain series by heart, like the ones you identified (trig, exp, geo). And you would also recognise them if one or a finite number of terms were missing (in this case the first term of the cosine series).
Then, if a series resembles one of these but there is an additional factor $n$ or $1/n$ in the coefficients, it is related to the first by integration or derivation (because the derivative of $x^n$ is $nx^{n-1}).$ If the factor is off by a constant, e.g., $n-2$ instead of $n,$ this means that the derivative applies to the series after multiplication by a constant power of $x.$ In our example the factors $1,3,5$ in the denominator are 'off by 2' with respect to the power of $a$ and the factorial in the denominator, e.g., we have a term
$$t(a)=-\frac{a^4}{3\cdot4!}$$
which suggests integration of $a^2,$ not $a^3$ or $a^4$. But notice that
$$a^2\frac{d}{da}\left(\frac1{a}t(a)\right)=-\frac{a^4}{4!}$$
so it becomes one of our standard series if we combine derivation with multiplication by a constant power of $a$ (before and after).
A: $$K(x)=4\sum_{n=0}^\infty \frac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)!}$$
Differentiate two times to get
$$K''(x)=4\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$$
Multiply by $x$ both sides:
$$xK''(x)=4\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}=\sin(x)$$
$$K''(x)=4\frac{\sin(x)}{x}$$
Integrate two times from $0$ to $x$:
$$K'(x)=4\mbox{Si}(x)$$
$$K(x)=4(x\mbox{Si}(x)+\cos(x)-1)$$
A: The general expression is $\frac{K}{4a^2} = \sum_{k=0}^\infty (-1)^k\frac{a^{2k}}{(2k+1) \cdot (2k)!}$
Hint: Note that
$\int (-1)^k\frac{a^{2k}}{(2k+1) \cdot (2k)!} da = (-1)^k\frac{a^{2k+1}}{ (2k)!}$
