\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Using the Maclaurin Expansion of \(e^x\) and setting \(x = 1\), we have \[ e = e^1 = \sum _{n = 0}^{\infty } \frac {1^n}{n!} = \sum _{n = 0}^{\infty } \frac {1}{n!}. \]
Hence, \begin {align*} \sum _{n = 1}^{\infty } \frac {n + 1}{n!} & = \sum _{n = 1}^{\infty } \frac {n}{n!} + \sum _{n = 1}^{\infty } \frac {1}{n!} \\ & = \sum _{n = 1}^{\infty } \frac {1}{(n - 1)!} + \sum _{n = 0}^{\infty } \frac {1}{n!} - \frac {1}{0!} \\ & = \sum _{n = 0}^{\infty } \frac {1}{n!} + \sum _{n = 0}^{\infty } \frac {1}{n!} - 1 \\ & = e + e - 1 \\ & = 2e-1. \end {align*}
We have as well \begin {align*} \sum _{n = 1}^{\infty } \frac {(n + 1)^2}{n!} & = \sum _{n = 1}^{\infty } \frac {n (n - 1) + 3n + 1}{n!} \\ & = \sum _{n = 1}^{\infty } \frac {n(n - 1)}{n!} + 3 \sum _{n = 1}^{\infty } \frac {n}{n!} + \sum _{n = 1}^{\infty } \frac {1}{n!} \\ & = \sum _{n = 2}^{\infty } \frac {1}{(n - 2)!} + 3 \sum _{n = 1}^{\infty } \frac {1}{(n - 1)!} + \sum _{n = 0}^{\infty } \frac {1}{n!} - 1 \\ & = 5\sum _{n = 0}^{\infty } \frac {1}{n!} - 1 \\ & = 5e - 1, \end {align*}
as desired.
We also have \begin {align*} \sum _{n = 1}^{\infty } \frac {(2n - 1)^3}{n!} & = \sum _{n = 1}^{\infty } \frac {8n^3 - 12n^2 + 6n - 1}{n!} \\ & = \sum _{n = 1}^{\infty } \frac {8n(n - 1)(n - 2) + 12 n (n - 1) + 2n - 1}{n!} \\ & = 8 \sum _{n = 3}^{\infty } \frac {1}{(n - 3)!} + 12 \sum _{n = 2}^{\infty }\frac {1}{(n - 2)!} + 2 \sum _{n = 1}^{\infty } \frac {1}{(n - 1)!} - \sum _{n = 0}^{\infty }\frac {1}{n!} + 1 \\ & = (8 + 12 + 2 - 1) \sum _{n = 0}^{\infty } \frac {1}{n!} + 1 \\ & = 21e + 1. \end {align*}
Using the Maclaurin Expansion of \(\ln (1 - x)\) and letting \(x = \frac {1}{2}\), we have \[ \ln 2 = - \ln \left (1 - \frac {1}{2}\right ) = \sum _{n = 1}^{\infty } \frac {\left (\frac {1}{2}\right )^n}{n} = \sum _{n = 1}^{\infty } \frac {2^{-n}}{n}. \]
Hence, \begin {align*} \sum _{n = 0}^{\infty } \frac {(n^2 + 1) 2^{-n}}{(n + 1)(n + 2)} & = \sum _{n = 0}^{\infty } \frac {[(n + 1)(n + 2) - 5(n + 1) + 2(n + 2)]2^{-n}}{(n + 1)(n + 2)} \\ & = \sum _{n = 0}^{\infty } 2^{-n} - 5\sum _{n = 0}^{\infty } \frac {2^{-n}}{n + 2} + 2 \sum _{n = 0}^{\infty } \frac {2^{-n}}{n + 1} \\ & = 2 - 5 \cdot 4 \sum _{n = 2}^{\infty } \frac {2^{-n}}{n} + 2 \cdot 2 \sum _{n = 1}^{\infty } \frac {2^{-n}}{n} \\ & = 2 - 20 (\ln 2 - \frac {1}{2}) +4 (\ln 2) \\ & = -16 \ln 2 + 12. \end {align*}