\(% 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}}\)
For each try, there is a probability of \(\frac {1}{n}\) of getting the correct key, and \(1 - \frac {1}{n}\) otherwise. Let \(X_1\) denote the number of attempts to open the door, we can see that \(X_1 \sim \Geometric \left (\frac {1}{n}\right )\), and hence using the formula for a geometric distribution, \[ \Expt (X_1) = n. \]
The way to consider the binomial expansion is as follows. First, note the probability mass function of \(X_1\) is \[ \Prob (X_1 = x) = \left (1 - \frac {1}{n}\right )^{x - 1} \cdot \frac {1}{n}, \] and hence the expectation is given by \begin {align*} \Expt (X_1) & = \sum _{x = 1}^{\infty } x \Prob (X_1 = x) \\ & = \sum _{x = 1}^{\infty } x \cdot \left (1 - \frac {1}{n}\right )^{x - 1} \cdot \frac {1}{n} \\ & = \frac {1}{n} \cdot \sum _{x = 1}^{\infty } x \cdot \left (1 - \frac {1}{n}\right )^{x - 1}. \end {align*}
Consider the binomial expansion of \((1 - q)^{-2}\). We have \begin {align*} (1 - q)^{-2} & = \sum _{t = 0}^{\infty } \frac {(-q)^t \cdot \prod _{r = 1}^{t} (-2 + 1 - t)}{t!} \\ & = \sum _{t = 0}^{\infty } \frac {(-1)^t q^t (-1)^t \prod _{r = 1}^{t} (1 + t)}{t!} \\ & = \sum _{t = 0}^{\infty } \frac {q^t (t + 1)!}{t!} \\ & = \sum _{t = 0}^{\infty } (t + 1) q^t. \end {align*}
Let \(q = 1 - \frac {1}{n}\). We can see \begin {align*} \Expt (X_1) & = \frac {1}{n} \cdot \sum _{x = 1}^{\infty } x \cdot \left (1 - \frac {1}{n}\right )^{x - 1} \\ & = \frac {1}{n} \cdot \sum _{x = 0}^{\infty } (x + 1) \cdot q^x \\ & = \frac {1}{n} \cdot (1 - q)^{-2} \\ & = \frac {1}{n} \cdot \left (\frac {1}{n}\right )^{-2} \\ & = n, \end {align*}
precisely what we had before.
Let \(X_2\) be the number of attempts to open the door in this case. Considering the probability mass function of \(X_2\), we have for \(x = 1, 2, \ldots , n\), that \begin {align*} \Prob (X_2 = x) & = \frac {n - 1}{n} \cdot \frac {n - 2}{n - 1} \cdots \frac {n - (x - 2) - 1}{n - (x - 2)} \cdot \frac {1}{n - (x - 1)} \\ & = \frac {(n - 1)! / (n - x)!}{n! / (n - x)!} \\ & = \frac {(n - 1)!}{n!} \\ & = \frac {1}{n}. \end {align*}
This shows that \(X_2\) follows a discrete uniform distribution on \(\{1, 2, \ldots , n\}\), i.e., \(X_2 \sim \Uniform (n)\).
Hence, \(\Expt (X_2) = \frac {n + 1}{2}\).
Let \(X_3\) be the number of attempts to open the door in this case. Considering the probability mass function of \(X_2\), we have for \(x = 1, 2, \ldots \), that \begin {align*} \Prob (X_3 = x) & = \frac {n - 1}{n} \cdot \frac {n}{n + 1} \cdots \frac {n + x - 3}{n + x - 2} \cdot \frac {1}{n + x - 1} \\ & = \frac {(n + x - 3)! / (n - 2)!}{(n + x - 1)! / (n - 1)!} \\ & = \frac {(n + x - 3)! (n - 1)!}{(n + x - 1)! (n - 2)!} \\ & = \frac {n - 1}{(n + x - 1) (n + x - 2)}, \end {align*}
which is precisely what is desired.
By partial fractions, we have \[ \Prob (X_3 = x) = (n - 1) \cdot \left (\frac {2}{n + x - 2} - \frac {1}{n + x - 1}\right ), \] and hence the expected number of attempts is \begin {align*} \Expt (X_3) & = \sum _{x = 1}^{\infty } (n - 1) \cdot x \cdot \left (\frac {1}{n + x - 2} - \frac {1}{n + x - 1}\right ) \\ & = (n - 1) \sum _{x = 1}^{\infty } x \left (\frac {1}{n + x - 2} - \frac {1}{n + x - 1}\right ). \end {align*}
We consider the partial sum of this infinite sum op to \(x = t\), and \begin {align*} \sum _{x = 1}^{t} x \left (\frac {1}{n + x - 2} - \frac {1}{n + x - 1}\right ) & = \sum _{x = 1}^{t} \frac {x}{n + x - 2} - \sum _{x = 1}^{t} \frac {x}{n + x - 1} \\ & = \sum _{x = 0}^{t - 1} \frac {x + 1}{n + x - 1} - \sum _{x = 1}^{t} \frac {x}{n + x - 1} \\ & = \frac {1}{n - 1} + \sum _{x = 1}^{t - 1} \frac {1}{n + x - 1} - \frac {t}{n + t - 1} \\ & = \sum _{x = 0}^{t - 1} \frac {1}{n + x - 1} - \frac {t}{n + t - 1} \\ & = \sum _{x = n - 1}^{n + t - 2} \frac {1}{x} - \frac {t}{n + t - 1}. \end {align*}
Hence, we have \begin {align*} \Expt (X_3) & = (n - 1) \sum _{x = 1}^{\infty } x \left (\frac {1}{n + x - 2} - \frac {1}{n + x - 1}\right ) \\ & = (n - 1) \lim _{t \to \infty } \left (\sum _{x = n - 1}^{n + t - 2} \frac {1}{x} - \frac {t}{n + t - 1}\right ) \\ & = (n - 1) \lim _{t \to \infty } \left (\sum _{x = 1}^{n + t - 2} \frac {1}{x} - \sum _{x = 1}^{n - 2} \frac {1}{x} - \frac {t}{n + t - 1}\right ) \\ & = (n - 1) \left (\sum _{x = 1}^{\infty } \frac {1}{x} - \sum _{x = 1}^{n - 2} \frac {1}{x} - 1\right ) \end {align*}
does not converge since the first term (harmonic sum) diverges, and the rest of the terms are finite.