\(% 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}}\)
We first look at the base case where \(n = 1\). \(S_1 = \frac {1}{1} = 1\), and \(2 \cdot \sqrt {1} - 1 = 1\), so \[ S_1 \leq 2 \cdot \sqrt {1} - 1 \] holds, and the original statement holds for when \(n = 1\).
Assume this holds for some \(n = k \in \NN \), i.e., \(S_k \leq 2 \sqrt {k} - 1\). We have \begin {align*} S_{k + 1} & = \sum _{r = 1}^{k + 1} \frac {1}{\sqrt {r}} \\ & = \sum _{r = 1}^{k} \frac {1}{\sqrt {r}} + \frac {1}{\sqrt {k + 1}} \\ & = S_k + \frac {1}{\sqrt {k + 1}} \\ & \leq 2 \sqrt {k} + \frac {1}{\sqrt {k + 1}} - 1. \end {align*}
We would like to show \[ 2\sqrt {k} + \frac {1}{\sqrt {k + 1}} \leq 2 \sqrt {k + 1}. \]
Notice that \begin {align*} 2\sqrt {k} + \frac {1}{\sqrt {k + 1}} \leq 2 \sqrt {k + 1} & \iff 2\sqrt {k(k + 1)} + 1 \leq 2 (k + 1) \\ & \iff 2\sqrt {k(k + 1)} \leq 2k + 1 \\ & \iff 4k(k + 1) \leq (2k + 1)^2 \\ & \iff 4k^2 + 4k \leq 4k^2 + 4k + 1 \\ & \iff 0 \leq 1, \end {align*}
which is true.
Hence, \[ S_{k + 1} \leq 2 \sqrt {k} + \frac {1}{\sqrt {k + 1}} - 1 \leq 2\sqrt {k + 1} - 1, \] which is precisely the statement for \(n = k + 1\).
The original statement holds for the base case where \(n = 0\), and assuming it holds for some \(n = k \in \NN \), it holds for \(n = k + 1\). Hence, by the principle of mathematical induction, the original statement holds for all \(n \in \NN \).
For \(k \geq 0\), we notice \begin {align*} (4k + 1) \sqrt {k + 1} > (4k + 3) \sqrt {k} & \iff (4k + 1)^2 (k + 1) > (4k + 3)^2 k \\ & \iff (16k^2 + 8k + 1) (k + 1) > (16k^2 + 24k + 9) k \\ & \iff 16k^3 + 8k^2 + k + 16k^2 + 8k + 1 > 16k^3 + 24k^2 + 9k \\ & \iff 1 > 0, \end {align*}
which is true.
We claim that \(C = \frac {3}{2}\) is the smallest number \(C\) which makes this true. We first show that \(C = \frac {3}{2}\) makes the statement true by induction. For the base case where \(n = 1\), \(S_1 = 1\), and \[ 2 \sqrt {1} + \frac {1}{2\sqrt {1}} - \frac {3}{2} = \frac {5}{2} - \frac {3}{2} = 1, \] and so this statement holds for \(n = 1\).
Now, assume that this statement holds for some \(n = k \in \NN \), i.e. \[ S_k \geq 2\sqrt {k} + \frac {1}{2\sqrt {k}} - C. \]
We have \begin {align*} S_{k + 1} & = S_k + \frac {1}{\sqrt {k + 1}} \\ & \geq 2\sqrt {k} + \frac {1}{2 \sqrt {k}} + \frac {1}{\sqrt {k + 1}} - C. \end {align*}
We would like to show that \[ 2\sqrt {k} + \frac {1}{2 \sqrt {k}} + \frac {1}{\sqrt {k + 1}} \geq 2\sqrt {k + 1} + \frac {1}{2\sqrt {k + 1}}. \]
Notice that \begin {align*} & \phantom {\iff } 2\sqrt {k} + \frac {1}{2 \sqrt {k}} + \frac {1}{\sqrt {k + 1}} \geq 2\sqrt {k + 1} + \frac {1}{2\sqrt {k + 1}} \\ & \iff 2\sqrt {k} + \frac {1}{2\sqrt {k}} \geq 2\sqrt {k + 1} - \frac {1}{2\sqrt {k + 1}} \\ & \iff \frac {4k + 1}{2\sqrt {k}} \geq \frac {4(k + 1) - 1}{2\sqrt {k + 1}} \\ & \iff (4k + 1) \sqrt {k + 1} \geq (4k + 3) \sqrt {k}, \end {align*}
which is implied by the proven inequality, and hence \[ S_{k + 1} \geq 2\sqrt {k} + \frac {1}{2 \sqrt {k}} + \frac {1}{\sqrt {k + 1}} - C \geq 2\sqrt {k + 1} + \frac {1}{2\sqrt {k + 1}} - C, \] which precisely proves the statement for \(n = k + 1\).
The claimed statement holds for the base case where \(n = 1\), and given it holds for some \(n = k \in \NN \), it holds for \(n = k + 1\). Hence, the statement holds for all \(n \in \NN \) when \(C = \frac {3}{2}\).
If \(C < \frac {3}{2}\), we have for \(n = 1\) \[ 2\sqrt {1} + \frac {1}{2 \sqrt {1}} - C > \frac {5}{2} - \frac {3}{2} = 1, \] but \[ S_1 = 1, \] so the statement does not hold for when \(n = 1\).
Hence, the smallest number \(C\) for the statement to be true is \(C = \frac {3}{2}\).