Prove that $\dfrac{4}{1!}+\dfrac{11}{2!}+\dfrac{22}{3!}+\dfrac{37}{4!}+\dfrac{56}{5!}+\cdots \cdots \cdots +\infty =6e-1$.

Hint: For finding the sum of the given series, we will first find the ${{n}^{th}}$ term ${{T}_{n}}$ of the series by finding ${{n}^{th}}$ term of the numerator and the denominator respectively. Then, to find the sum ${{S}_{n}}$ we will use ${{S}_{n}}=\sum{{{T}_{n}}}$. We will use following formula or properties:
(i) Sum of n terms of an AP is equal to ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ where a is the first term of AP and d is the common difference.
$\begin{align}
  & \left( ii \right)\sum\limits_{n=1}^{\infty }{\dfrac{1}{n!}}=e-1 \\
 & \left( iii \right)\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-1 \right)!}}=e=\sum\limits_{n=1}^{\infty }{\dfrac{n}{n!}} \\
 & \left( iv \right)\sum\limits_{n=1}^{\infty }{\dfrac{{{n}^{2}}}{\left( n-1 \right)!}}=2e \\
\end{align}$

Complete step by step answer:
Here we are given the series as $\dfrac{4}{1!}+\dfrac{11}{2!}+\dfrac{22}{3!}+\dfrac{37}{4!}+\dfrac{56}{5!}+\cdots \cdots \cdots +\infty $ we need to prove it to be equal to $6e-1$.
For this let us first find the ${{n}^{th}}$ term of the given series represented by $6e-1$.
We will find ${{n}^{th}}$ term for numerator and denominator separately.
For numerator, we have the series as $4,11,22,37,56,\ldots \ldots $.
We need to find its ${{n}^{th}}$ term. So let us suppose it to ${{a}_{n}}$. We can write the sum as,
\[S=4+11+22+37+56+\ldots \ldots +{{a}_{n-1}}+{{a}_{n}}+0\cdots \cdots \cdots \left( 1 \right)\]S
Similarly, we can write this term again as,
\[S=0+4+11+22+37+56+\ldots \ldots +{{a}_{n-1}}+{{a}_{n}}\cdots \cdots \cdots \left( 2 \right)\]
Now subtracting (2) from (1),
\[\begin{align}
  & S=4+11+22+37+56+\cdots +{{a}_{n-1}}+{{a}_{n}}+0 \\
 & S=0+4+11+22+37+\cdots +{{a}_{n-2}}+{{a}_{n-1}}+{{a}_{n}} \\
 & \begin{matrix}
   - & - & - & - & - & - & - & - & - & - & - & - \\
\end{matrix} \\
 & \overline{0=4+7+11+15+19+\cdots +\left( {{a}_{n}}-{{a}_{n-1}} \right)-{{a}_{n}}} \\
\end{align}\]
Taking ${{a}_{n}}$ to the other side we get,
$\begin{align}
  & {{a}_{n}}=4+7+11+15+19+\ldots \ldots +\left( {{a}_{n}}-{{a}_{n-1}} \right) \\
 & \Rightarrow {{a}_{n}}=4+\left( 7+11+15+19+\ldots \ldots +\left( {{a}_{n}}-{{a}_{n-1}} \right) \right) \\
\end{align}$
We can see that $11-7=4=15-11$ so, $7+11+15+19+\ldots \ldots +\left( {{a}_{n}}-{{a}_{n-1}} \right)$ is in arithmetic progression having (n-1) terms. First term as 7 and common difference as 4.
We know that, sum of n terms of an AP is given by ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.
Now, for (n-1) terms with a = 7 and d = 4 we get,
$\begin{align}
  & 7+11+15+19+\ldots \ldots +\left( {{a}_{n}}-{{a}_{n-1}} \right)=\dfrac{\left( n-1 \right)}{2}\left( 2\left( 7 \right)+\left( \left( n-1 \right)-1 \right)4 \right) \\
 & \Rightarrow \dfrac{\left( n-1 \right)}{2}\left( 14+\left( n-2 \right)4 \right) \\
\end{align}$
So ${{a}_{n}}$ becomes, ${{a}_{n}}=4+\dfrac{\left( n-1 \right)}{2}\left( 14+\left( n-2 \right)4 \right)$.
Simplifying it we get, ${{a}_{n}}=4+\dfrac{\left( n-1 \right)}{2}\left( 14+4n-8 \right)\Rightarrow {{a}_{n}}=4+\dfrac{\left( n-1 \right)}{2}\left( 6+4n \right)$.
Dividing 2 by (6+4x) we get, ${{a}_{n}}=4+\left( n-1 \right)\left( 3+2n \right)$.
Simplifying it we get, ${{a}_{n}}=4+\left( 3n+2{{n}^{2}}-3-2n \right)$.
Adding and subtracting like terms we get, ${{a}_{n}}=4+n+2{{n}^{2}}-3\Rightarrow {{a}_{n}}=2{{n}^{2}}+n+1$.
Hence ${{n}^{th}}$ term of the numerator of the given series is $2{{n}^{2}}+n+1$.
Also the denominator of the series is 1!, 2!, 3!, . . . . . . . . . .
We know that, its ${{n}^{th}}$ term will be n!
Hence we can say that ${{n}^{th}}$ term of the series is $\dfrac{2{{n}^{2}}+n+1}{n!}$ so, ${{T}_{n}}=\dfrac{2{{n}^{2}}+n+1}{n!}$.
We need to find sum of the series which is equal to $\sum{{{T}_{n}}}$.
So we need to find ${{S}_{n}}$ given as ${{S}_{n}}=\sum{{{T}_{n}}}=\sum{\dfrac{2{{n}^{2}}+n+1}{n!}}\Rightarrow {{S}_{n}}=\sum{\left( \dfrac{2{{n}^{2}}}{n!}+\dfrac{n}{n!}+\dfrac{1}{n!} \right)}$.
We know that, we can separate terms for summation. So we get, ${{S}_{n}}=\sum{\dfrac{2{{n}^{2}}}{n!}+\sum{\dfrac{n}{n!}}+\sum{\dfrac{1}{n!}}}$.
Also we can take constant out of the summation so we get, \[{{S}_{n}}=2\sum{\dfrac{{{n}^{2}}}{n!}+\sum{\dfrac{n}{n!}}+\sum{\dfrac{1}{n!}}}\].
We know that $\sum{\dfrac{{{n}^{2}}}{n!}}$ is equal to 2e, $\sum{\dfrac{n}{n!}}$ is equal to e and $\sum{\dfrac{1}{n!}}$ is equal to e-1. So putting in the values we get,
$\begin{align}
  & {{S}_{n}}=2\left( 2e \right)+\left( e \right)+\left( e-1 \right) \\
 & \Rightarrow {{S}_{n}}=4e+e+e-1 \\
 & \Rightarrow {{S}_{n}}=6e-1 \\
\end{align}$
Hence the sum of the given series is 6e-1.
Hence proved.

Note: Students should note that, we have used ${{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{x}{2!}+\dfrac{x}{3!}+\ldots \ldots $ to find the values of $\sum{\dfrac{1}{n!}}$,$\sum{\dfrac{n}{n!}}$ and $\sum{\dfrac{{{n}^{2}}}{n!}}$. They should keep in mind all the formula for solving this sum. Take care of the signs while solving the sum. Keep in mind formulas for summation of the same series.