Odd Series

$x=\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+\frac{4}{9!}+\ldots$

or, $2x=\frac{3-1}{3!}+\frac{5-1}{5!}+\frac{7-1}{7!}+\frac{9-1}{9!}+\ldots$

or, $2x=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}+\frac{1}{8!}-\frac{1}{9!}+\ldots$

or, $x=\frac{1}{2e}$ .

Now, $\left(\lceil{1+\frac{1}{x}}\rceil\right)! = 7! = \boxed{5040}$ .

Consider the Maclaurin series of $\sinh t$ :

$\sinh t = \sum_{n=0}^\infty \frac{t^{2n+1}}{(2n+1)!}.$

Substituting $t = \sqrt{u}$ and diving both sides by $\sqrt{v}$ , we obtain

$\frac{\sinh \sqrt{u}}{\sqrt{u}} = \sum_{n=0}^\infty \frac{u^n}{(2n+1)!}.$

Differentiating wrt u at u = 1 gives us

$\left. \frac{d}{du} \frac{\sinh \sqrt{u}}{\sqrt{u}} \right|_{u=1} = \left. \sum_{n=0}^\infty \frac{nu^{n-1}}{(2n+1)!} \right|_{u=1} = \sum_{n=0}^\infty \frac{n}{(2n+1)!}.$

The RHS is our desired sum (since the n = 0 term is 0), which leaves us to handle the LHS. A quick calculation yields

$\left. \frac{d}{du} \frac{\sinh \sqrt{u}}{\sqrt{u}} \right|_{u=1} = \frac{\cosh 1 - \sinh 1}{2} = \frac{1}{2e}.$

Thus,

$x = \sum_{n=0}^\infty \frac{n}{(2n+1)!} = \boxed{\frac{1}{2e}}$

and so $\left( \left\lceil 1+\dfrac{1}{x} \right\rceil \right) ! = 7! = \boxed{5040}$ .

First, let's find $x$ . It is the sum

$x = \sum_{n=1}^{\infty} \frac{n}{(2n+1)!}$

Now, if you remember your calculus lessons:

$\sinh(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$

Which is easy to differentiate:

$\frac{\mathrm{d} }{\mathrm{d} x} \sinh(x) = \sum_{n=0}^{\infty} \frac{(2n+1) x^{2n}}{(2n+1)!} = \cosh(x)$

And therefore

$\cosh(1) = \sum_{n= 0 or 1}^{\infty} \frac{2n}{(2n+1)} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = 2x + \sinh(1)$

So $2x = e + e^{-1} - (e - e^{-1}) = e^{-1}$ or $1/x = 2e$ . Therefore

$\left( \left\lceil 1 + \frac{1}{x} \right\rceil \right)! = \left( \left\lceil 1 + 2e \right\rceil \right)! = 7! = 5040$