Key Concepts and Formulas

• The definition of the mathematical constant $e$: $e = \sum_{n=0}^{\infty} \frac{1}{n!}$.
• The power series expansion of $e^x$: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$. This implies $e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}$.
• The hyperbolic cosine and sine series:
  • $\cosh(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$
  • $\sinh(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$ For $x=1$, we have:
  • $\cosh(1) = \frac{e + e^{-1}}{2} = \sum_{n=0}^{\infty} \frac{1}{(2n)!}$
  • $\sinh(1) = \frac{e - e^{-1}}{2} = \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \sum_{n=1}^{\infty} \frac{1}{(2n-1)!}$
• Algebraic manipulation of series and index shifting.

Step-by-Step Solution

1. Decompose the Given Series: The given series is \sum_\limits{n=0}^{\infty} \frac{\mathrm{n}^{3}((2 \mathrm{n}) !)+(2 \mathrm{n}-1)(\mathrm{n} !)}{(\mathrm{n} !)((2 \mathrm{n}) !)}. We can split this into two fractions: $\sum_{n=0}^{\infty} \left( \frac{n^3 (2n)!}{n! (2n)!} + \frac{(2n-1) n!}{n! (2n)!} \right) = \sum_{n=0}^{\infty} \left( \frac{n^3}{n!} + \frac{2n-1}{(2n)!} \right)$ This decomposition allows us to analyze each part separately.

2. Evaluate the First Part: $\sum_{n=0}^{\infty} \frac{n^3}{n!}$ We need to express $n^3$ in terms of falling factorials to simplify the series. Let $n^3 = A n(n-1)(n-2) + B n(n-1) + C n$. For $n=1$: $1^3 = A(0) + B(0) + C(1) \implies C=1$. For $n=2$: $2^3 = A(0) + B(2)(1) + C(2) \implies 8 = 2B + 2(1) \implies 2B = 6 \implies B=3$. For $n=3$: $3^3 = A(3)(2)(1) + B(3)(2) + C(3) \implies 27 = 6A + 6(3) + 3(1) \implies 27 = 6A + 18 + 3 \implies 27 = 6A + 21 \implies 6A = 6 \implies A=1$. So, $n^3 = n(n-1)(n-2) + 3n(n-1) + n$. Now, substitute this back into the series: $\sum_{n=0}^{\infty} \frac{n^3}{n!} = \sum_{n=0}^{\infty} \frac{n(n-1)(n-2) + 3n(n-1) + n}{n!}$ $= \sum_{n=0}^{\infty} \frac{n(n-1)(n-2)}{n!} + 3\sum_{n=0}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=0}^{\infty} \frac{n}{n!}$ Let's evaluate each sum:

   • $\sum_{n=0}^{\infty} \frac{n(n-1)(n-2)}{n!} = \sum_{n=3}^{\infty} \frac{n(n-1)(n-2)}{n!} = \sum_{n=3}^{\infty} \frac{1}{(n-3)!}$. Let $k = n-3$. When $n=3$, $k=0$. $\sum_{k=0}^{\infty} \frac{1}{k!} = e$
   • $3\sum_{n=0}^{\infty} \frac{n(n-1)}{n!} = 3\sum_{n=2}^{\infty} \frac{n(n-1)}{n!} = 3\sum_{n=2}^{\infty} \frac{1}{(n-2)!}$. Let $k = n-2$. When $n=2$, $k=0$. $3\sum_{k=0}^{\infty} \frac{1}{k!} = 3e$
   • $\sum_{n=0}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{1}{(n-1)!}$. Let $k = n-1$. When $n=1$, $k=0$. $\sum_{k=0}^{\infty} \frac{1}{k!} = e$ Therefore, $\sum_{n=0}^{\infty} \frac{n^3}{n!} = e + 3e + e = 5e$.

3. Evaluate the Second Part: $\sum_{n=0}^{\infty} \frac{2n-1}{(2n)!}$ We can split this sum into two: $\sum_{n=0}^{\infty} \frac{2n-1}{(2n)!} = \sum_{n=0}^{\infty} \frac{2n}{(2n)!} - \sum_{n=0}^{\infty} \frac{1}{(2n)!}$ Consider the first sum: $\sum_{n=0}^{\infty} \frac{2n}{(2n)!}$. For $n=0$, the term is $\frac{0}{0!} = 0$. So, we start from $n=1$: $\sum_{n=1}^{\infty} \frac{2n}{(2n)!} = \sum_{n=1}^{\infty} \frac{1}{(2n-1)!}$ This is the series for $\sinh(1)$: $\sum_{n=1}^{\infty} \frac{1}{(2n-1)!} = \sinh(1) = \frac{e - e^{-1}}{2}$ Consider the second sum: $\sum_{n=0}^{\infty} \frac{1}{(2n)!}$. This is the series for $\cosh(1)$: $\sum_{n=0}^{\infty} \frac{1}{(2n)!} = \cosh(1) = \frac{e + e^{-1}}{2}$ So, the second part of the original series is: $\sum_{n=0}^{\infty} \frac{2n-1}{(2n)!} = \frac{e - e^{-1}}{2} - \frac{e + e^{-1}}{2} = \frac{e - e^{-1} - e - e^{-1}}{2} = \frac{-2e^{-1}}{2} = -e^{-1} = -\frac{1}{e}$

4. Combine the Results and Identify Coefficients: The total sum is the sum of the two parts: $\sum_{n=0}^{\infty} \frac{n^3}{n!} + \sum_{n=0}^{\infty} \frac{2n-1}{(2n)!} = 5e + \left(-\frac{1}{e}\right) = 5e - \frac{1}{e}$ We are given that the sum is equal to $ae + \frac{b}{e} + c$, where $a, b, c \in \mathbb{Z}$. By comparing $5e - \frac{1}{e}$ with $ae + \frac{b}{e} + c$, we can identify the coefficients: $a = 5$ $b = -1$ $c = 0$

5. Calculate the Final Expression: We need to find the value of $a^2 - b + c$. $a^2 - b + c = (5)^2 - (-1) + 0 = 25 + 1 + 0 = 26$

Common Mistakes & Tips

• Algebraic Errors: Carefully expand and simplify expressions involving factorials and powers.
• Index Shifting: When changing the index of summation (e.g., from $n$ to $k$), ensure the starting and ending values of the index are adjusted correctly.
• Recognizing Series: Be familiar with the power series expansions of $e^x$, $\cosh(x)$, and $\sinh(x)$, as they are frequently used in such problems.
• Falling Factorials: Expressing polynomials in terms of falling factorials ($n(n-1)(n-2)...$) simplifies series involving $n!$ in the denominator.

Summary

The problem required evaluating an infinite series by decomposing it into two parts. The first part, $\sum_{n=0}^{\infty} \frac{n^3}{n!}$, was simplified by expressing $n^3$ in terms of falling factorials, leading to a sum involving $e$. The second part, $\sum_{n=0}^{\infty} \frac{2n-1}{(2n)!}$, was evaluated using the series expansions of $\sinh(1)$ and $\cosh(1)$. By combining these results and comparing with the given form $ae + \frac{b}{e} + c$, we found the integer coefficients $a=5$, $b=-1$, and $c=0$. Finally, $a^2 - b + c$ was calculated.

The final answer is $\boxed{26}$.