CMIMC 2017

First of all recall that:

$\large \displaystyle e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$

Differentiating with respect to $x$ and multiplying by $x$ :

$\large \displaystyle xe^x = \sum_{k=0}^{\infty} \frac{kx^k}{k!}$

Making $x=1$ on both equations:

$\color{#20A900} \boxed{ \large \displaystyle e = \sum_{k=0}^{\infty} \frac{1}{k!} = \sum_{k=0}^{\infty} \frac{k}{k!} }$

Now, to the problem:

$\large \displaystyle S = \sum_{k=3}^{\infty} \frac{k}{\left \lfloor \frac{k}{2} \right \rfloor ! }$

$\large \displaystyle S = 3 + \frac{4}{2!} + \frac{5}{2!} + \frac{6}{3!} + \frac{7}{3!} +...$

$\large \displaystyle S = 3 + \sum_{k=2}^{\infty} \frac{2k + 2k+1}{k!}$

$\large \displaystyle S = 3 + 4\sum_{k=2}^{\infty} \frac{k}{k!} + \sum_{k=2}^{\infty} \frac{1}{k!}$

$\large \displaystyle S = 3 + 4\left [ \sum_{k=0}^{\infty} \frac{k}{k!} - 1 \right ] + \left [ \sum_{k=0}^{\infty} \frac{1}{k!} - 2 \right ]$

$\large \displaystyle S = 3 + 4(e-1) + (e-2)$

$\color{#20A900} \boxed{\large \displaystyle S = 5e-3}$

Thus:

$\large \displaystyle \color{#3D99F6} A=5, B = 3, \boxed{\large \displaystyle A+B=8}$

$\begin{aligned} S & = \sum_{k=3}^\infty \frac k{\left \lfloor \frac k2 \right \rfloor !} \\ & = {\color{#3D99F6}\frac 3{1!}} + {\color{#D61F06}\frac 4{2!}} + {\color{#3D99F6}\frac 5{2!}} + {\color{#D61F06}\frac 6{3!}}+ {\color{#3D99F6}\frac 7{3!}} + {\color{#D61F06}\frac 8{4!}} + \cdots \\ & = {\color{#3D99F6} \sum_{k=1}^\infty \frac {2k+1}{k!}} + {\color{#D61F06} \sum_{k=2}^\infty \frac {2k}{k!}} \\ & = {\color{#3D99F6} 2\sum_{k=1}^\infty \frac 1{(k-1)!} +\sum_{k=1}^\infty \frac 1{k!}} + {\color{#D61F06} 2\sum_{k=2}^\infty \frac 1{(k-1)!}} \\ & = 2\sum_{\color{#D61F06}k=0}^\infty \frac 1{k!} + \sum_{\color{#D61F06}k=0}^\infty \frac 1{k!} {\color{#D61F06}- 1} + 2 \left(\sum_{\color{#D61F06}k=0}^\infty \frac 1{k!} {\color{#D61F06}- 1}\right) \\ & = 5 \sum _{k=0}^\infty \frac 1{k!} - 3 \\ & = 5e - 3 \end{aligned}$

$\implies A + B = 5+3 = \boxed{8}$