Key Concepts and Formulas

• Maclaurin Series for $e^x$ and $e^{-x}$:
  • $e^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
  • $e^{-x} = \sum_{k=0}^\infty \frac{(-x)^k}{k!} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$
• Sum and Difference of $e$ and $e^{-1}$:
  • $e + e^{-1} = 2 \sum_{k=0}^\infty \frac{1}{(2k)!} = 2 \left(1 + \frac{1}{2!} + \frac{1}{4!} + \dots\right)$
  • $e - e^{-1} = 2 \sum_{k=0}^\infty \frac{1}{(2k+1)!} = 2 \left(1 + \frac{1}{3!} + \frac{1}{5!} + \dots\right)$
• Manipulation of Series Terms: Expressing a polynomial in the numerator in terms of falling factorials (e.g., $r(r-1)$, $r$) allows for simplification when divided by $r!$.

Step-by-Step Solution

Step 1: Analyze the General Term and Prepare for Substitution The given series is $S = \sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}}$. The presence of $(2n+1)!$ in the denominator suggests we should aim to express the numerator in a form that can be simplified using factorial properties. Let $T_n = {{{n^2} + 6n + 10} \over {(2n + 1)!}}$. To work with the factorial, we introduce a substitution. Let $r = 2n+1$. From $r = 2n+1$, we get $n = \frac{r-1}{2}$. Since $n$ starts from $1$, the smallest value for $r$ is $2(1)+1 = 3$. As $n$ takes integer values $1, 2, 3, \dots$, $r$ takes odd integer values $3, 5, 7, \dots$.

Step 2: Substitute and Simplify the Numerator in terms of $r$ Substitute $n = \frac{r-1}{2}$ into the numerator $n^2 + 6n + 10$: $n^2 + 6n + 10 = \left(\frac{r-1}{2}\right)^2 + 6\left(\frac{r-1}{2}\right) + 10$ $= \frac{(r-1)^2}{4} + 3(r-1) + 10$ $= \frac{r^2 - 2r + 1}{4} + 3r - 3 + 10$ $= \frac{r^2 - 2r + 1}{4} + 3r + 7$ To combine these terms, we find a common denominator: $= \frac{r^2 - 2r + 1 + 4(3r + 7)}{4}$ $= \frac{r^2 - 2r + 1 + 12r + 28}{4}$ $= \frac{r^2 + 10r + 29}{4}$ So, the general term in terms of $r$ is: $T_r = \frac{\frac{r^2 + 10r + 29}{4}}{r!} = \frac{r^2 + 10r + 29}{4 \cdot r!}$

Step 3: Decompose the Numerator into Falling Factorials Our goal is to rewrite the numerator $r^2 + 10r + 29$ in a form that simplifies nicely with $r!$. We express it in terms of falling factorials: $r(r-1)$, $r$, and a constant. Let $r^2 + 10r + 29 = A \cdot r(r-1) + B \cdot r + C$. Expanding the right side gives $Ar^2 - Ar + Br + C = Ar^2 + (B-A)r + C$. Comparing coefficients with $r^2 + 10r + 29$:

• Coefficient of $r^2$: $A = 1$.
• Coefficient of $r$: $B-A = 10$. Since $A=1$, $B-1=10 \implies B = 11$.
• Constant term: $C = 29$. Thus, $r^2 + 10r + 29 = 1 \cdot r(r-1) + 11 \cdot r + 29$.

Step 4: Rewrite the General Term using the Decomposed Numerator Now substitute this back into the expression for $T_r$: $T_r = \frac{r(r-1) + 11r + 29}{4 \cdot r!}$ We can split this into three terms: $T_r = \frac{1}{4} \left( \frac{r(r-1)}{r!} + \frac{11r}{r!} + \frac{29}{r!} \right)$ Simplify each term:

• $\frac{r(r-1)}{r!} = \frac{r(r-1)}{r(r-1)(r-2)!} = \frac{1}{(r-2)!}$ (for $r \ge 2$)
• $\frac{11r}{r!} = \frac{11r}{r(r-1)!} = \frac{11}{(r-1)!}$ (for $r \ge 1$)
• $\frac{29}{r!} = \frac{29}{r!}$

So, $T_r = \frac{1}{4} \left( \frac{1}{(r-2)!} + \frac{11}{(r-1)!} + \frac{29}{r!} \right)$.

Step 5: Express the Series in terms of $r$ and sum The original series is $\sum_{n=1}^\infty T_n$. With the substitution $r=2n+1$, the sum becomes a sum over odd integers $r \ge 3$: $S = \sum_{r \in \{3, 5, 7, \dots\}} \frac{1}{4} \left( \frac{1}{(r-2)!} + \frac{11}{(r-1)!} + \frac{29}{r!} \right)$ Let's expand the sum for the first few terms of $r$: For $r=3$: $\frac{1}{4} \left( \frac{1}{1!} + \frac{11}{2!} + \frac{29}{3!} \right)$ For $r=5$: $\frac{1}{4} \left( \frac{1}{3!} + \frac{11}{4!} + \frac{29}{5!} \right)$ For $r=7$: $\frac{1}{4} \left( \frac{1}{5!} + \frac{11}{6!} + \frac{29}{7!} \right)$ And so on.

Let's separate the sum based on the factorial terms: $4S = \sum_{r \in \{3, 5, 7, \dots\}} \frac{1}{(r-2)!} + 11 \sum_{r \in \{3, 5, 7, \dots\}} \frac{1}{(r-1)!} + 29 \sum_{r \in \{3, 5, 7, \dots\}} \frac{1}{r!}$

Consider the first sum: $\sum_{r \in \{3, 5, 7, \dots\}} \frac{1}{(r-2)!}$. Let $k = r-2$. When $r=3$, $k=1$. When $r=5$, $k=3$. When $r=7$, $k=5$. So the terms are $\frac{1}{1!}, \frac{1}{3!}, \frac{1}{5!}, \dots$. This sum is $\sum_{j=0}^\infty \frac{1}{(2j+1)!} = \frac{e - e^{-1}}{2}$.

Consider the second sum: $11 \sum_{r \in \{3, 5, 7, \dots\}} \frac{1}{(r-1)!}$. Let $k = r-1$. When $r=3$, $k=2$. When $r=5$, $k=4$. When $r=7$, $k=6$. So the terms are $\frac{1}{2!}, \frac{1}{4!}, \frac{1}{6!}, \dots$. This sum is $11 \sum_{j=1}^\infty \frac{1}{(2j)!} = 11 \left( \sum_{j=0}^\infty \frac{1}{(2j)!} - \frac{1}{0!} \right) = 11 \left( \frac{e + e^{-1}}{2} - 1 \right)$.

Consider the third sum: $29 \sum_{r \in \{3, 5, 7, \dots\}} \frac{1}{r!}$. The terms are $\frac{1}{3!}, \frac{1}{5!}, \frac{1}{7!}, \dots$. This sum is $29 \left( \sum_{j=0}^\infty \frac{1}{(2j+1)!} - \frac{1}{1!} \right) = 29 \left( \frac{e - e^{-1}}{2} - 1 \right)$.

Step 6: Combine the Sums and Solve for S Now, substitute these back into the expression for $4S$: $4S = \left(\frac{e - e^{-1}}{2}\right) + 11 \left(\frac{e + e^{-1}}{2} - 1\right) + 29 \left(\frac{e - e^{-1}}{2} - 1\right)$ $4S = \frac{e - e^{-1}}{2} + \frac{11(e + e^{-1})}{2} - 11 + \frac{29(e - e^{-1})}{2} - 29$ $4S = \frac{e - e^{-1} + 11e + 11e^{-1} + 29e - 29e^{-1}}{2} - 40$ Combine the terms with $e$: $e + 11e + 29e = 41e$. Combine the terms with $e^{-1}$: $-e^{-1} + 11e^{-1} - 29e^{-1} = -19e^{-1}$. $4S = \frac{41e - 19e^{-1}}{2} - 40$ $S = \frac{1}{4} \left( \frac{41e - 19e^{-1}}{2} - 40 \right)$ $S = \frac{41e - 19e^{-1}}{8} - \frac{40}{4}$ $S = \frac{41}{8}e - \frac{19}{8}e^{-1} - 10$

Common Mistakes & Tips

• Incorrectly Handling the Range of $r$: Ensure that when substituting $n$ with $r$, the starting value and progression of $r$ (odd numbers $\ge 3$) are correctly used to define the summation limits or the terms included in the series.
• Algebraic Errors in Numerator Simplification: Be meticulous when expanding and combining terms after substituting $n = \frac{r-1}{2}$. A small mistake here can lead to a completely different result.
• Forgetting the $-1$ terms: When using the sums for even and odd factorials, remember that $\sum_{k=0}^\infty \frac{1}{(2k)!} = \frac{e+e^{-1}}{2}$ includes the $\frac{1}{0!}=1$ term, and $\sum_{k=0}^\infty \frac{1}{(2k+1)!} = \frac{e-e^{-1}}{2}$ includes the $\frac{1}{1!}=1$ term. Adjustments are needed if the series starts from a different index.

Summary

The problem involves summing an infinite series with a polynomial numerator and a factorial in the denominator. The strategy is to transform the general term by substituting the denominator's structure ($2n+1$) with a new variable ($r$) to obtain a simpler factorial expression. The polynomial numerator is then decomposed into falling factorials, allowing the general term to be split into parts that match known Maclaurin series expansions for $e^x$ and $e^{-x}$. By carefully summing these parts and performing algebraic manipulations, the final sum of the series is obtained.

The final answer is $\boxed{{\frac{41}{8}e + \frac{19}{8}{e^{ - 1}} - 10}}$.