\(% 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}}\)
\begin {align*} \sum _{k = 1}^{N} \frac {k + 1}{k!} \cdot x^k & = \sum _{k = 1}^{N} \frac {k}{k!} \cdot x^k + \sum _{k = 1}^{N} \frac {x^k}{k!} \\ & = \sum _{k = 1}^{N} \frac {1}{(k - 1)!} \cdot x^k + \sum _{k = 0}^{N} \frac {x^k}{k!} - \frac {x^0}{0!} \\ & = \sum _{k = 0}^{N - 1} \frac {1}{k!} \cdot x^{k + 1} + \sum _{k = 0}^{N} \frac {x^k}{k!} - 1 \\ & = x \sum _{k = 0}^{N - 1} \frac {x^k}{k!} + \sum _{k = 0}^{N} \frac {x^k}{k!} - 1. \end {align*}
We let \(N \to \infty \). Using the Maclaurin Expansion for \(e^x\), we have \[ \sum _{k = 0}^{\infty } \frac {x^k}{k!} = e^x, \] and hence \[ \sum _{k = 1}^{\infty } \frac {k + 1}{k!} \cdot x^k = x e^x + e^x - 1 = (x + 1) e^x - 1. \]
We have \(Y \sim \Poisson (n)\). Let \(X_k\) be the outcome of a \(k\)-sided die, i.e. \(X_k \sim \Uniform (k)\). WE must have \(1 \leq X_k \leq k\). The random variable \(D\) can be defined as \[ D = \begin {cases} 0, & Y = 0, \\ X_k, & Y = k. \end {cases} \]
For \(d \geq 1\), we have \begin {align*} \Prob (D = d) & = \sum _{k = d}^{\infty } \Prob (X_k = d, Y = k) \\ & = \sum _{k = d}^{\infty } \Prob (X_k = d) \Prob (Y = k) \\ & = \sum _{k = d}^{\infty } \frac {1}{k} \cdot e^{-n} \cdot \frac {n^k}{k!} \\ & = \sum _{k = d}^{\infty } \left (\frac {1}{k} \cdot \frac {n^k}{k!} \cdot e^{-n}\right ). \end {align*}
Hence, \begin {align*} \Expt (D) & = \sum _{d = 0}^{\infty } d \Prob (D = d) \\ & = \sum _{d = 1}^{\infty } d \Prob (D = d) \\ & = \sum _{d = 1}^{\infty } \left [d \sum _{k = d}^{\infty } \left (\frac {1}{k} \cdot \frac {n^k}{k!} \cdot e^{-n}\right )\right ]. \end {align*}
This summation is for \[ d \cdot \left (\frac {1}{k} \cdot \frac {n^k}{k!} \cdot e^{-n}\right ) \] over the set \begin {align*} (d, k) & \in \{(n, m) \mid n \geq 1, m \geq n\} \\ & = \{(n, m) \mid 1 \leq n \leq m\} \\ & = \{(n, m) \mid m \geq 1, n \leq m\}. \end {align*}
Therefore, \begin {align*} \Expt (D) & = \sum _{d = 1}^{\infty } \left [d \sum _{k = d}^{\infty } \left (\frac {1}{k} \cdot \frac {n^k}{k!} \cdot e^{-n}\right )\right ] \\ & = \sum _{(d, k) \in \{(n, m) \mid n \geq 1, m \geq n\}} d \cdot \left (\frac {1}{k} \cdot \frac {n^k}{k!} \cdot e^{-n}\right ) \\ & = \sum _{(d, k) \in \{(n, m) \mid m \geq 1, n \leq m\}} d \cdot \left (\frac {1}{k} \cdot \frac {n^k}{k!} \cdot e^{-n}\right ) \\ & = \sum _{k = 1}^{\infty } \sum _{d = 1}^{k} d \cdot \left (\frac {1}{k} \cdot \frac {n^k}{k!} \cdot e^{-n}\right ) \\ & = \sum _{k = 1}^{\infty } \left [\frac {1}{k} \cdot \frac {n^k}{k!} \cdot e^{-n} \cdot \sum _{d = 1}^{k} d\right ]. \end {align*}
\begin {align*} \Expt (D) & = \sum _{k = 1}^{\infty } \left [\frac {1}{k} \cdot \frac {n^k}{k!} \cdot e^{-n} \cdot \sum _{d = 1}^{k} d\right ] \\ & = \sum _{k = 1}^{\infty } \left [\frac {1}{k} \cdot \frac {n^k}{k!} \cdot e^{-n} \cdot \frac {k (k + 1)}{2}\right ] \\ & = \frac {e^{-n}}{2} \sum _{k = 1}^{\infty } \frac {n^k (k + 1)}{k!} \\ & = \frac {e^{-n}}{2} \left [(n + 1) \cdot e^n - 1\right ] \\ & = \frac {1}{2} \left [e^{-n} \cdot (n + 1) \cdot e^{n} - e^{-n}\right ] \\ & = \frac {1}{2} \left [(n + 1) - e^{-n}\right ] \end {align*}
as desired.
\(X_k \sim \Poisson (k)\) for \(k = 1, 2, \cdots , n\). Let \(Y_n\) be the outcome of an \(n\)-sided die, i.e. \(Y_n \sim \Uniform (n)\). Therefore, \(Z = X_k\) if \(Y_n = k\).
For \(z \geq 1\), we have \begin {align*} \Prob (Z = z) & = \sum _{k = 1}^{n} \Prob (X_k = z, Y_n = k) \\ & = \sum _{k = 1}^{n} \Prob (X_k = z) \Prob (Y_n = k) \\ & = \frac {1}{n} \cdot \sum _{k = 1}^{n} e^{-k} \cdot \frac {k^z}{z!} \\ & = \frac {1}{nz!} \sum _{k = 1}^{n} e^{-k} k^{z}. \end {align*}
Hence, \begin {align*} \Expt (Z) & = \sum _{z = 0}^{\infty } z \Prob (Z = z) \\ & = \sum _{z = 1}^{\infty } z \Prob (Z = z) \\ & = \sum _{z = 1}^{\infty } \left [\frac {1}{n (z - 1)!} \cdot \sum _{k = 1}^{n} e^{-k} \cdot k^Z\right ] \\ & = \frac {1}{n} \sum _{z = 1}^{\infty } \left [\frac {1}{(z - 1)!} \sum _{k = 1}^{n} e^{-k} \cdot k^z\right ] \\ & = \frac {1}{n} \sum _{z = 1}^{\infty } \sum _{k = 1}^{n} \left (\frac {1}{(z - 1)!} \cdot e^{-k} \cdot k^z\right ) \\ & = \frac {1}{n} \sum _{k = 1}^{n} \sum _{z = 1}^{\infty } \left (\frac {1}{(z - 1)!} \cdot e^{-k} \cdot k^z\right ) \\ & = \frac {1}{n} \sum _{k = 1}^{n} \left [e^{-k} \cdot k \cdot \sum _{z = 1}^{\infty } \frac {k^{z - 1}}{(z - 1)!}\right ] \\ & = \frac {1}{n} \sum _{k = 1}^{n} \left [e^{-k} \cdot k \cdot \sum _{z = 0}^{\infty } \frac {k^z}{z!}\right ] \\ & = \frac {1}{n} \sum _{k = 1}^{n} \left [e^{-k} \cdot k \cdot e^{k}\right ] \\ & = \frac {1}{n} \sum _{k = 1}^{n} k \\ & = \frac {1}{n} \cdot \frac {n (n + 1)}{2} \\ & = \frac {n + 1}{2}. \end {align*}
Therefore, subtracting gives us \begin {align*} \Expt (Z) - \Expt (D) & = \frac {n + 1}{2} - \frac {1}{2} \cdot \left (n + 1 - e^{-n}\right ) \\ & = \frac {1}{2} e^{-n} \\ & > 0. \end {align*}
Therefore, \(\Expt (Z) > \Expt (D)\) as desired.