Sequence and Series 2

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

Dividing by $x$ on both sides:

$\large \displaystyle \frac{e^x}{x} = \sum_{k=0}^{\infty} \frac{x^{k-1}}{k!}$

Differentiating with respect to $x$ on both sides:

$\large \displaystyle \frac{e^x(x-1)}{x^2} = \sum_{k=0}^{\infty} \frac{(k-1)x^{k-2}}{k!}$

Multiplying by $x$ on both sides:

$\large \displaystyle e^x - \frac{e^x}{x} = \sum_{k=0}^{\infty} \frac{(k-1)x^{k-1}}{k!}$

Differentiating with respect to $x$ on both sides:

$\large \displaystyle e^x - \frac{e^x(x-1)}{x^2} = \sum_{k=0}^{\infty} \frac{(k-1)^2 x^{k-2}}{k!}$

Making $x=1$ :

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

Defining $k = n+1$ :

$\large \displaystyle e = \sum_{n=-1}^{\infty} \frac{n^2}{(n+1)!}$

$\large \displaystyle e = \frac{(-1)^2}{0!} + \frac{0^2}{1!} + \sum_{n=1}^{\infty} \frac{n^2}{(n+1)!}$

$\large \displaystyle \sum_{n=1}^{\infty} \frac{n^2}{(n+1)!} = e-1$

$\large \displaystyle \color{#20A900} \boxed{ \mathfrak{B} = e-1 }$

Thus:

$\large \displaystyle \color{#3D99F6} \boxed{\lceil \mathfrak{B} \rceil = 2 }$