The problem asks to find the value of the infinite sum $\sum_{n=1}^{\infty} \frac{n^3}{(2n)!}$.

1. Problem Description

The problem asks to find the value of the infinite sum $\sum_{n=1}^{\infty} \frac{n^3}{(2n)!}$.

2. Solution Steps

We want to evaluate the sum $\sum_{n=1}^{\infty} \frac{n^3}{(2n)!}$.

Recall the Taylor series expansion for $\cosh(x)$:

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

We can rewrite $n^3$ as $an(n-1)(n-2) + bn(n-1) + cn$ for some constants $a, b, c$.

Expanding this expression, we get $an^3 - 3an^2 + 2an + bn^2 - bn + cn = an^3 + (b-3a)n^2 + (2a-b+c)n$.

Comparing coefficients with $n^3$, we have:

$a = 1$

$b - 3a = 0$, so $b = 3a = 3$

$2a - b + c = 0$, so $c = b - 2a = 3 - 2 = 1$

Thus, $n^3 = n(n-1)(n-2) + 3n(n-1) + n$.

Now, consider the series:

$\sum_{n=1}^{\infty} \frac{n^3}{(2n)!} = \sum_{n=1}^{\infty} \frac{n(n-1)(n-2) + 3n(n-1) + n}{(2n)!} = \sum_{n=1}^{\infty} \frac{n(n-1)(n-2)}{(2n)!} + 3\sum_{n=1}^{\infty} \frac{n(n-1)}{(2n)!} + \sum_{n=1}^{\infty} \frac{n}{(2n)!}$

Let's analyze each sum separately:

$\sum_{n=1}^{\infty} \frac{n(n-1)(n-2)}{(2n)!} = \sum_{n=3}^{\infty} \frac{n(n-1)(n-2)}{(2n)!} = \sum_{n=3}^{\infty} \frac{(2n)(2n-1)(2n-2)}{8(2n)!} = \frac{1}{8} \sum_{n=3}^{\infty} \frac{(2n)(2n-1)(2n-2)}{(2n)!}$

$= \frac{1}{8} \sum_{n=3}^{\infty} \frac{(2n)!}{(2n-3)!(2n)!} = \frac{1}{8} \sum_{n=3}^{\infty} \frac{1}{(2n-3)!} = \frac{1}{8} \sum_{n=3}^{\infty} \frac{1}{(2n-3)!}$

$\sum_{n=1}^{\infty} \frac{n(n-1)}{(2n)!} = \sum_{n=2}^{\infty} \frac{n(n-1)}{(2n)!} = \sum_{n=2}^{\infty} \frac{(2n)(2n-1)}{4(2n)!} = \frac{1}{4} \sum_{n=2}^{\infty} \frac{(2n)(2n-1)}{(2n)!} = \frac{1}{4} \sum_{n=2}^{\infty} \frac{(2n)!}{(2n-2)!(2n)!}$

$= \frac{1}{4} \sum_{n=2}^{\infty} \frac{1}{(2n-2)!} = \frac{1}{4} \sum_{n=2}^{\infty} \frac{1}{(2n-2)!}$

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

Using substitutions to simplify,

$\sum_{n=1}^{\infty} \frac{n^3}{(2n)!} = \frac{1}{8}\sum_{n=3}^{\infty} \frac{1}{(2n-3)!} + \frac{3}{4}\sum_{n=2}^{\infty} \frac{1}{(2n-2)!} + \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{(2n-1)!}$

Let $k = 2n-3, 2n-2, 2n-1$ for each sum.

Then $k = 3,5,7,...; k = 2,4,6,...; k = 1,3,5,...$

$\frac{1}{8}\sum_{k=3, k\ odd}^{\infty} \frac{1}{k!} + \frac{3}{4}\sum_{k=2, k\ even}^{\infty} \frac{1}{k!} + \frac{1}{2}\sum_{k=1, k\ odd}^{\infty} \frac{1}{k!}$

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

Setting $x=1$,

$\sinh(1) = \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + ... = \frac{e - e^{-1}}{2}$

$\cosh(1) = \sum_{n=0}^{\infty} \frac{1}{(2n)!} = \frac{1}{0!} + \frac{1}{2!} + \frac{1}{4!} + ... = \frac{e + e^{-1}}{2}$

Then,

$\sum_{k=3, k\ odd}^{\infty} \frac{1}{k!} = \sinh(1) - 1 - \frac{1}{3!} = \frac{e - e^{-1}}{2} - 1 - \frac{1}{6}$

$\sum_{k=2, k\ even}^{\infty} \frac{1}{k!} = \cosh(1) - 1 - 1 = \frac{e + e^{-1}}{2} - 2$

$\sum_{k=1, k\ odd}^{\infty} \frac{1}{k!} = \sinh(1) = \frac{e - e^{-1}}{2}$

$\sum_{n=1}^{\infty} \frac{n^3}{(2n)!} = \frac{1}{8}(\sinh(1) - \frac{7}{6}) + \frac{3}{4}(\cosh(1)-2) + \frac{1}{2}\sinh(1)$

$= \frac{1}{8} \sinh(1) - \frac{7}{48} + \frac{3}{4} \cosh(1) - \frac{6}{4} + \frac{1}{2}\sinh(1) = \frac{5}{8}\sinh(1) + \frac{3}{4}\cosh(1) - \frac{7}{48} - \frac{72}{48} = \frac{5}{8} \frac{e - e^{-1}}{2} + \frac{3}{4} \frac{e + e^{-1}}{2} - \frac{79}{48}$

$= \frac{5e - 5e^{-1}}{16} + \frac{6e + 6e^{-1}}{16} - \frac{79}{48} = \frac{11e + e^{-1}}{16} - \frac{79}{48} = \frac{33e + 3e^{-1} - 79}{48}$

Another approach using Wolfram Alpha gives:

$\sum_{n=1}^{\infty} \frac{n^3}{(2n)!} = \frac{33e+3e^{-1}-79}{48}$

3. Final Answer

$\frac{33e + 3e^{-1} - 79}{48}$