Key Concepts and Formulas

• Taylor Series for $e^x$: The exponential function $e^x$ has a well-known Taylor series expansion around $x=0$: $e^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
• Series for $e$ and $e^{-1}$: Substituting $x=1$ and $x=-1$ into the Taylor series for $e^x$ yields: $e = \sum_{k=0}^\infty \frac{1}{k!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ $e^{-1} = \frac{1}{e} = \sum_{k=0}^\infty \frac{(-1)^k}{k!} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots$
• Series for Even and Odd Factorials: By adding and subtracting the series for $e$ and $e^{-1}$, we can derive series involving only even or odd factorials: $\sum_{k=0}^\infty \frac{1}{(2k)!} = \frac{e + e^{-1}}{2}$ $\sum_{k=0}^\infty \frac{1}{(2k+1)!} = \frac{e - e^{-1}}{2}$
• Numerator Decomposition Strategy: For series of the form $\sum \frac{P(n)}{(kn)!}$, where $P(n)$ is a polynomial, we decompose $P(n)$ into a linear combination of terms involving falling factorials related to the factorial in the denominator. For $(2n)!$, we aim for terms like $(2n)(2n-1)$, $(2n)$, and a constant.

Step-by-Step Solution

The problem asks us to evaluate the infinite sum $S = \sum\limits_{n = 1}^\infty {{{2{n^2} + 3n + 4} \over {(2n)!}}}$.

Step 1: Decompose the Numerator Our first step is to rewrite the polynomial in the numerator, $2n^2 + 3n + 4$, in a form that will simplify when divided by $(2n)!$. We aim to express it as a linear combination of terms that can be easily cancelled by the factorial. Specifically, we look for constants $A, B, C$ such that: $2n^2 + 3n + 4 = A(2n)(2n-1) + B(2n) + C$ Expanding the right side gives: $A(4n^2 - 2n) + 2Bn + C = 4An^2 + (-2A + 2B)n + C$ By comparing the coefficients of $n^2$, $n$, and the constant term on both sides, we get: \begin{itemize} \item Coefficient of $n^2$: $4A = 2 \implies A = \frac{1}{2}$ \item Coefficient of $n$: $-2A + 2B = 3$. Substituting $A = \frac{1}{2}$, we get $-2(\frac{1}{2}) + 2B = 3 \implies -1 + 2B = 3 \implies 2B = 4 \implies B = 2$. \item Constant term: $C = 4$. \end{itemize} Thus, the numerator can be written as $2n^2 + 3n + 4 = \frac{1}{2}(2n)(2n-1) + 2(2n) + 4$.

Step 2: Split the Series Substitute the decomposed numerator back into the sum: $S = \sum\limits_{n = 1}^\infty \frac{\frac{1}{2}(2n)(2n-1) + 2(2n) + 4}{(2n)!}$ We can split this into three separate infinite series: $S = \sum\limits_{n = 1}^\infty \frac{\frac{1}{2}(2n)(2n-1)}{(2n)!} + \sum\limits_{n = 1}^\infty \frac{2(2n)}{(2n)!} + \sum\limits_{n = 1}^\infty \frac{4}{(2n)!}$ Now, we simplify each term using the property $(2n)! = (2n)(2n-1)(2n-2)!$ or $(2n)! = (2n)(2n-1)!$: \begin{itemize} \item First term: $\frac{\frac{1}{2}(2n)(2n-1)}{(2n)!} = \frac{\frac{1}{2}(2n)(2n-1)}{(2n)(2n-1)(2n-2)!} = \frac{1}{2(2n-2)!}$ \item Second term: $\frac{2(2n)}{(2n)!} = \frac{2(2n)}{(2n)(2n-1)!} = \frac{2}{(2n-1)!}$ \item Third term: $\frac{4}{(2n)!}$ (remains as is for now) \end{itemize} So the sum becomes: $S = \sum\limits_{n = 1}^\infty \frac{1}{2(2n-2)!} + \sum\limits_{n = 1}^\infty \frac{2}{(2n-1)!} + \sum\limits_{n = 1}^\infty \frac{4}{(2n)!}$

Step 3: Evaluate Each Sub-Series Let's evaluate each of these three series separately.

Sub-series 1: $S_1 = \sum\limits_{n = 1}^\infty \frac{1}{2(2n-2)!}$ $S_1 = \frac{1}{2} \sum\limits_{n = 1}^\infty \frac{1}{(2n-2)!}$ Let $k = n-1$. When $n=1$, $k=0$. When $n \to \infty$, $k \to \infty$. The term $2n-2$ becomes $2(k+1)-2 = 2k$. $S_1 = \frac{1}{2} \sum\limits_{k = 0}^\infty \frac{1}{(2k)!}$ Using the formula $\sum_{k=0}^\infty \frac{1}{(2k)!} = \frac{e + e^{-1}}{2}$: $S_1 = \frac{1}{2} \left( \frac{e + e^{-1}}{2} \right) = \frac{e + e^{-1}}{4}$

Sub-series 2: $S_2 = \sum\limits_{n = 1}^\infty \frac{2}{(2n-1)!}$ $S_2 = 2 \sum\limits_{n = 1}^\infty \frac{1}{(2n-1)!}$ Let $k = n-1$. When $n=1$, $k=0$. The term $2n-1$ becomes $2(k+1)-1 = 2k+1$. $S_2 = 2 \sum\limits_{k = 0}^\infty \frac{1}{(2k+1)!}$ Using the formula $\sum_{k=0}^\infty \frac{1}{(2k+1)!} = \frac{e - e^{-1}}{2}$: $S_2 = 2 \left( \frac{e - e^{-1}}{2} \right) = e - e^{-1}$

Sub-series 3: $S_3 = \sum\limits_{n = 1}^\infty \frac{4}{(2n)!}$ $S_3 = 4 \sum\limits_{n = 1}^\infty \frac{1}{(2n)!}$ The series $\sum\limits_{n = 1}^\infty \frac{1}{(2n)!}$ includes terms $\frac{1}{2!}, \frac{1}{4!}, \frac{1}{6!}, \dots$. We know that $\sum_{k=0}^\infty \frac{1}{(2k)!} = \frac{1}{0!} + \frac{1}{2!} + \frac{1}{4!} + \dots = \frac{e + e^{-1}}{2}$. So, $\sum\limits_{n = 1}^\infty \frac{1}{(2n)!} = \left( \sum\limits_{k=0}^\infty \frac{1}{(2k)!} \right) - \frac{1}{0!} = \frac{e + e^{-1}}{2} - 1$. Therefore, $S_3 = 4 \left( \frac{e + e^{-1}}{2} - 1 \right) = 2(e + e^{-1}) - 4$

Step 4: Combine the Sub-Series The total sum $S$ is the sum of $S_1, S_2,$ and $S_3$: $S = S_1 + S_2 + S_3$ $S = \frac{e + e^{-1}}{4} + (e - e^{-1}) + (2(e + e^{-1}) - 4)$ Group terms involving $e$ and $e^{-1}$: $S = \left(\frac{1}{4}e + e + 2e\right) + \left(\frac{1}{4}e^{-1} - e^{-1} + 2e^{-1}\right) - 4$ $S = \left(\frac{1}{4} + 1 + 2\right)e + \left(\frac{1}{4} - 1 + 2\right)e^{-1} - 4$ $S = \left(\frac{1 + 4 + 8}{4}\right)e + \left(\frac{1 - 4 + 8}{4}\right)e^{-1} - 4$ $S = \frac{13}{4}e + \frac{5}{4}e^{-1} - 4$ This can be written as: $S = \frac{13e}{4} + \frac{5}{4e} - 4$

Common Mistakes & Tips

• Index of Summation: Carefully check the starting index of the given series ($n=1$ in this case). This affects whether the $k=0$ term (e.g., $1/0!$) is included in the standard formulas.
• Factorial Simplification: Ensure correct application of factorial properties like $(2n)! = (2n)(2n-1)(2n-2)!$.
• Algebraic Errors: Double-check arithmetic when combining coefficients of $e$ and $e^{-1}$ from different series.
• Numerator Decomposition: The method of expressing the polynomial numerator in terms of falling factorials is crucial for simplifying series with factorials in the denominator.

Summary The problem involves evaluating an infinite series with a polynomial in the numerator and a factorial in the denominator. The strategy is to decompose the numerator into terms that cancel with the factorial, allowing us to split the series into simpler components. These components are then related to the known series expansions of $e$ and $e^{-1}$, particularly the sums involving even and odd factorials. By carefully performing the decomposition, simplification, and summation, we arrive at the final value of the series.

The final answer is $\boxed{\text{B}}$.