Calculate: $\frac{1}{2! \cdot 2} + \frac{1}{4! \cdot 4} + \frac{1}{6! \cdot 8} + \frac{1}{8! \cdot 16} + ...$ Task from an old exam:

Calculate (express without an infinite sum): $$\frac{1}{2! \cdot 2} +
\frac{1}{4! \cdot 4} + \frac{1}{6! \cdot 8} + \frac{1}{8! \cdot 16} +
...$$

I think this means on the way to the solution, we are allowed to use the sum symbol but the final result may not be in a sum symbol. Else I don't see another way of solving this task.
In sum, the thing above would be:
$$\sum_{n=1}^{\infty} \frac{1}{(2n)!\cdot 2^{n}}$$
And now we need to do something so the sum symbol is eliminated but I have no idea what it could be... 
What about seeing $$\frac{1}{(2n)!\cdot 2^{n}}$$ as a function and using taylor series on this? But no it would be too hard to derivate something like that and with factorial.. and I don't think it makes sense saying "Hey, let's replace this sum symbol with a function and say this thing is a function now!" 
But what else can I do in a situation like this? If you are doing something complicated, please do explain me. I have big troubles in understanding things.
 A: Note that the sum is the even terms of $f(x)=\sum \frac{x^n}{n!}$ evaluated at $x=\sqrt{2}/2$ (but ignoring the $0$th term).  This is the Taylor series for $e^x$.  To eliminate the odd terms, we take $(f(x)+f(-x))/2=\frac{e^x+e^{-x}}{2}$.  To remove the $0$th term, we subtract $f(0)$.  Putting everything together, the sum is $\frac{e^{\sqrt 2/2}+e^{-\sqrt 2/2}}{2}-1$.
A: taylor series of $\cosh x$ is
$$\cosh x=\sum_{n=0}^{\infty }\frac{x^{2n}}{(2n)!}$$
$$\cosh x=1+\sum_{n=1}^{\infty }\frac{x^{2n}}{(2n)!}$$
let $x=\frac{1}{\sqrt{2}}$
$$\cosh \frac{1}{\sqrt{2}}=1+\sum_{n=1}^{\infty }\frac{1}{2^{n}(2n)!}$$
A: In this case, we are trying to get a grasp on the function $\displaystyle\sum_{i=1}^\infty \frac{2^{-n}}{(2n)!}$.
Now, if this does remind you of anything, it is of the cosine series, where the coefficients are also of denominator $(2n)!$ Therefore, we would like something such that $(x^{2n}) = 2^{-n}$, so that we can write the expression as $\cos(x) - 1$ (because the cosine series begins from zero, we need to subtract $1$ to take care of that index) and be done. This $x$ is seen to be $i 2^{-0.5}$. Therefore, we would be able to write $\cos(2^{-0.5}i) - 1$. Using the result that $\cos(ix) = \frac{e^{x} + e^{-x}}{2}$, we see the answer as $\dfrac{e^{(2^{-0.5})} + e^{(-2^{-0.5})}}{2} -1$.
As a heuristic, one should always keep Taylor series at hand, and especially  when the coefficient are recognizable, the cosine and sine series expansions will come in useful, as it did so in this case.
