Does it converge?

I don't know why it's so under-rated but it's a really good problem and the closed form is the following -

$\displaystyle \sqrt{e} + \sqrt{\frac{e\pi}{2}}erf(\frac{1}{\sqrt{2}})$ (though erf will not be considered analytic)

So, Archit Boobna consider adding this in the question and asking $\displaystyle a + b + c + d + e..$ (I hope you are getting me) rather than a decimal form.

We know $\displaystyle cos(x\pi) = {(-1)}^{x}$ for integers $\displaystyle x$ .

The sum becomes -

$\displaystyle \sum_{x=1}^{\infty}{\frac{{2}^{\frac{x}{2}}{(\frac{\pi}{2})}^{\frac{{(-1)}^{x}-1}{4}}\Gamma(\frac{x}{2}+1)}{\Gamma(x+1)}}$

Now we will change into odd and even,

$\displaystyle \text{EVEN} - \sum_{k=1}^{\infty}{\frac{{2}^{k}*k!}{(2k)!}}$

Let's solve this first! We will use Beta function here.

$\displaystyle \sum_{k=1}^{\infty}{\frac{{2}^{k}*(k-1)!*k!}{(k-1)!*(2k)!}}$

$\displaystyle \sum_{k=1}^{\infty}{\frac{{2}^{k}\int_{0}^{1}{{t}^{k-1}{(1-t)}^{k}}}{(k-1)!}}$

$\displaystyle \int_{0}^{1}{\frac{1}{t}(\sum_{k=1}^{\infty}{\frac{{2t(1-t)}^{k}}{(k-1)!}}}$

$\displaystyle \int_{0}^{1}{2(1-t){e}^{2t(1-t)}}$

which can easily be solved by completing the square and then using the fact that $\displaystyle \text{erf}(a) = \int_{0}^{x}{{e}^{-{a}^{2}}}$

and the value we get is - $\displaystyle \sqrt{\frac{e\pi}{2}}\text{erf}(\frac{1}{\sqrt{2}})$ [ If anybody doesn't get this step, you can ask in the comments ]

Now, $\displaystyle \text{ODD} - \sum_{j=0}^{\infty}{\sqrt{\frac{2}{\pi}}\frac{{2}^{\frac{2j+1}{2}}\Gamma(j+1+\frac{1}{2})}{\Gamma(2j+2)}}$

Now, we will use one 'formula' whose source is this , proof of which is left as an exercise to the reader(but still you can ask in the comments)

$\displaystyle\Gamma(n + 1/2) = \frac{(2n)!\sqrt{\pi}}{{4}^{n}n!}$

And substituting this for $\displaystyle n=j+1$ ,

$\displaystyle \sum_{j=0}^{\infty}{\sqrt{\frac{2}{\pi}}\frac{{2}^{\frac{2j+1}{2}}\frac{(2j+2)!\sqrt{\pi}}{{2}^{2j+2}(j+1)!}}{(2j+1)!}}$

which simplifies to-

$\displaystyle \sum_{j=0}^{\infty}{\frac{1}{{2}^{j}*j!}}$

And hence,

$\displaystyle \sqrt{e}$

Finally adding the 2 answers, we get our final answer -

$\displaystyle \boxed{\sqrt{e} + \sqrt{\frac{e\pi}{2}}\text{erf}(\frac{1}{\sqrt{2}})}$