Infinite Factorial Summation with a twist - Part 2

It is given that the numerators follow a quadratic polynomial. Let that polynomial be $f(n)$ .

We proceed to find the unique quadratic polynomial $f(n)$ that satisfies $f(1)=2,f(2)=12,f(3)=28$ using the method of finite differences and then verify if the polynomial obtained matches the numerators of the subsequent terms given in the series. We start by constructing difference table for $f$ as follows:

$\begin{array}{|c|c|c|c|}\hline n&f(n)&D_1(n)&D_2(n)\\\hline 1&2&10&6\\2&12&16\\3&28\\\hline\end{array}$

Using the reconstruction formula, we have,

$f(n)=f(1)+\sum_{k=1}^2\left[\frac{D_k(1)}{k!}\prod_{j=1}^k(n-j)\right]=3n^2+n-2$

Checking for $f(4),f(5)$ shows that the polynomial indeed matches the numerators of the subsequent terms and the given sum to be computed is,

$\mathcal{S}=\sum_{k=1}^\infty\frac{f(k)}{k!}$

The sum converges absolutely (verified by ratio test).

We now compute the sum by breaking it into fragments that can be computed using series for $e$ .

$\begin{array}{c}\mathcal{S}=\sum_{k=1}^\infty\frac{f(k)}{k!}=\sum_{k=1}^\infty\frac{3k^2+k-2}{k!}\\\implies\mathcal{S}=3\left(\sum_{k=1}^\infty\frac{k}{(k-1)!}\right)+\sum_{k=1}^\infty\frac{1}{(k-1)!}-2\left(\sum_{k=1}^\infty\frac{1}{k!}\right)\\\implies\mathcal{S}=3\left(\sum_{k=1}^\infty\frac{(k-1)+1}{(k-1)!}\right)+\sum_{k=0}^\infty\frac{1}{k!}-2\left(\sum_{k=1}^\infty\frac{1}{k!}\right)\\\implies\mathcal{S}=3\left(\sum_{k=1}^\infty\frac{k-1}{(k-1)!}+\sum_{k=1}^\infty\frac{1}{(k-1)!}\right)+\sum_{k=0}^\infty\frac{1}{k!}-2\left(\sum_{k=1}^\infty\frac{1}{k!}\right)\\\implies\mathcal{S}=3\left(0+\sum_{k=2}^\infty\frac{k-1}{(k-1)!}+\sum_{k=1}^\infty\frac{1}{(k-1)!}\right)+\sum_{k=0}^\infty\frac{1}{k!}-2\left(\sum_{k=1}^\infty\frac{1}{k!}\right)\\\implies\mathcal{S}=3\left(\sum_{k=2}^\infty\frac{1}{(k-2)!}+\sum_{k=1}^\infty\frac{1}{(k-1)!}\right)+\sum_{k=0}^\infty\frac{1}{k!}-2\left(\sum_{k=1}^\infty\frac{1}{k!}\right)\\\implies\mathcal{S}=3\left(\sum_{k=0}^\infty\frac{1}{k!}+\sum_{k=0}^\infty\frac{1}{k!}\right)+\sum_{k=0}^\infty\frac{1}{k!}-2\left(\sum_{k=1}^\infty\frac{1}{k!}\right)\end{array}$

So, we now use the Maclaurin series for $e^x$ which converges $\forall~x$ . For the case of $x=1$ , we have the following series for $e$ :

$e=\sum_{k=0}^\infty\frac{1}{k!}=1+\sum_{k=1}^\infty\frac{1}{k!}$

The sum $\mathcal{S}$ evaluates as,

$\mathcal{S}=3(e+e)+e-2(e-1)=2+5e$

Comparing values, we get $A=2,B=5$ .

Hence, the answer is $A+B=2+5=\boxed{7}$ .

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Note: The "sum-breaking" done in 2nd and 4th lines of the calculation is valid because the sums obtained after "breaking" are also convergent sums (can be easily verified using convergence tests).