Identify the general term

let's say $T_{n}$ represents the $n^{th}$ term of the given series.

So, $\large{T_{n}=\frac{2n^{2}+n+1}{n!}}$

$\large{T_{n}=\frac{2n}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}}$

$\large{T_{n}=\frac{2}{(n-2)!}+\frac{3}{(n-1)!}+\frac{1}{n!}}$

let $S$ represents the sum of the series.

$\displaystyle S=2.\sum_{n=2}^\infty \left( \frac{1}{(n-2)!} \right)+3.\sum_{n=1}^\infty \left( \frac{1}{(n-1)!} \right)+\sum_{n=1}^\infty \left( \frac{1}{n!} \right)$

$\large{S=2e+3e+e-1=6e-1}$

$S=15.3096$

therefore, $\large \lfloor S \rfloor =15$

ENJOY!

Here we can easily Notice That Numerator has Second order difference Constant so it's general term is an polynomial of degree $2$ . So By putting initial Value We can find it.

$But$ There is another $way$ to find $'general$ $term'$ .

Let $Numerator$ be:

${ S }_{ n }=\quad 4+11+22+37+56\quad ...........{ T }_{ n-1 }+{ T }_{ n }$ .

Now Using Difference Method shift every term by one step:

${ S }_{ n }=\quad 4+11+22+37+56\quad ...........{ T }_{ n-1 }+{ T }_{ n }\quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow (1)\\ \\ { S }_{ n }=\quad \quad \quad \quad 4+11+22+37+56\quad ...........{ \quad T }_{ n-1 }\quad +\quad { T }_{ n }\quad \quad \longrightarrow (2)$ .

Now subtract the two equations and then we get:

$0=\quad 4+7+11+15+19\quad .......................({ T }_{ n }-{ T }_{ n-1 })\quad -{ \quad T }_{ n }$ .

This can be written as:

${ \quad T }_{ n }=4+7+11+15+19\quad .......................({ T }_{ n }-{ T }_{ n-1 })$ .

Now From $Second$ $term$ this sequence is an $AP$ with $'n-1'$ terms So:

${ T }_{ n }=\quad 4\quad +\quad \frac { (n-1)(14+4(n-2)) }{ 2 }$ .

${ T }_{ n }=\quad 2{ n }^{ 2 }+n+1$ .

For $Denominator$ general term is:

${ D }_{ n }=\quad n!$ .

So $P=\sum _{ 1 }^{ \infty }{ \frac { 2{ n }^{ 2 }+n+1 }{ n! } }$ .

Now for Further Calculation See nicely explain @Sandeep Bhardwaj 's Solution.

Let's say $T_{n}$ represents the $n^{th}$ term of the given series.

It is easily observed that $T_{n} =\frac{4+(7+11+15.....4n+3)}{(n+1)!}$ . Applying formula for ap with first and last term.

$T_{n} =\frac{4+\frac{n}{2}(7+4n+3)}{(n+1)!}$ $\displaystyle P=\sum ^\infty _{n=0} \frac{2n^2+5n+4}{(n+1)!}$ $\displaystyle P=\sum ^\infty _{n=0} \frac{(2n^2+2n)+(3n+3)+1}{(n+1)!}$ $\displaystyle P=\sum ^\infty _{n=1} \frac{2}{(n-1)!} +\sum ^\infty _{n=0} \frac{3}{n!}+\sum ^\infty _{n=0} \frac {1}{(n+1)!}$

$P=2e+3e+e-1 =6e-1$ $P=15.304$

Its floor i.e $\boxed {15}$ is the required answer.