\(% 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 have \begin {align*} \text {RHS} & = \frac {r+1}{r} \left (\frac {1}{\binom {n+r-1}{r}} - \frac {1}{\binom {n+r}{r}}\right ) \\ & = \frac {r+1}{r} \left (\frac {r! (n-1)!}{(n+r-1)!} - \frac {r! n!}{(n+r)!}\right ) \\ & = \frac {r+1}{r} \left (\frac {r! (n-1)! (n+r)}{(n+r)!} - \frac {r! (n-1)! n}{(n+r)!}\right ) \\ & = \frac {r+1}{r} \cdot \frac {r! (n-1)! (n+r) - r! (n-1)! n}{(n+r)!} \\ & = \frac {r+1}{r} \cdot \frac {r! (n-1)! r}{(n+r)!} \\ & = \frac {(r+1)! (n-1)! }{(n+r)!} \\ & = \binom {n+r}{r+1} \\ & = \text {LHS} \end {align*}
as desired.
Therefore, \begin {align*} \sum _{n = 1}^{\infty } \frac {1}{\binom {n+r}{r+1}} & = \sum _{n = 1}^{\infty } \frac {r+1}{r} \left (\frac {1}{\binom {n+r-1}{r}} - \frac {1}{\binom {n+r}{r}}\right ) \\ & = \frac {r+1}{r} \sum _{n = 1}^{\infty } \left (\frac {1}{\binom {n+r-1}{r}} - \frac {1}{\binom {n+r}{r}}\right ) \\ & = \frac {r+1}{r} \left [\sum _{n = 0}^{\infty } \frac {1}{\binom {n + r}{r}} - \sum _{n = 1}^{\infty } \frac {1}{\binom {n + r}{r}}\right ] \\ & = \frac {r+1}{r} \frac {1}{\binom {0+r}{r}} \\ & = \frac {r+1}{r}, \end {align*}
assuming the sum converges.
When \(r = 2\), we have \[ \sum _{n = 1}^{\infty } \frac {1}{\binom {n+2}{3}} = \frac {3}{2}. \]
When \(n = 1\), \(\frac {1}{\binom {1 + 2}{3}} = \frac {1}{1} = 1\).
Therefore, \[ \sum _{n = 2}^{\infty } \frac {1}{\binom {n+2}{3}} = \frac {1}{2} \] as desired.
Notice that \begin {align*} \frac {3!}{n^3} < \frac {1}{\binom {n+1}{3}} & \iff \frac {3!}{n^3} < \frac {3!}{(n+1)n(n-1)} \\ & \iff n^3 > (n+1)n(n-1) \\ & \iff n^3 > n(n^2 - 1) \\ & \iff n^3 > n^3 - n \\ & \iff n > 0, \end {align*}
which is true.
Also, notice that \begin {align*} \frac {20}{\binom {n+1}{3}} - \frac {1}{\binom {n+2}{5}} < \frac {5!}{n^3} & \iff \frac {5!}{(n+1)(n)(n-1)} - \frac {5!}{(n+2)(n+1)(n)(n-1)(n-2)} < \frac {5!}{n^3} \\ & \iff \frac {(n+2)(n-2) - 1}{(n+2)(n+1)(n)(n-1)(n-2)} < \frac {1}{n^3} \\ & \iff (n^2 - 5)n^3 < (n^2 - 4)(n^2 - 1)n \\ & \iff n^5 - 5n^3 < n^5 - 5n^3 + 4n \\ & \iff 4n > 0, \end {align*}
which is true.
Therefore, we have that \begin {align*} \sum _{n = 3}^{\infty } \frac {3!}{n^3} & < \sum _{n = 3}^{\infty } \frac {1}{\binom {n+1}{3}} \\ & = \sum _{n = 2}^{\infty } \frac {1}{\binom {n+2}{3}} \\ & = \frac {1}{2}, \end {align*}
and therefore \(\sum _{n = 3}^{\infty } \frac {1}{n^3} < \frac {1}{12}\), and \(\sum _{n = 1}^{\infty } \frac {1}{n^3} < 1 + \frac {1}{8} + \frac {1}{12} = \frac {29}{24} = \frac {116}{96}.\)
On the other hand, we have \begin {align*} \sum _{n = 3}^{\infty } \frac {5!}{n^3} & < \sum _{n = 3}^{\infty } \left [\frac {20}{\binom {n+1}{3}} - \frac {1}{\binom {n+2}{5}}\right ] \\ & = 20 \sum _{n = 2}^{\infty } \frac {1}{\binom {n+2}{3}} - \sum _{n = 1}^{\infty } \frac {1}{\binom {n+4}{5}} \\ & = 20 \cdot \frac {1}{2} - \frac {5}{4} \\ & = 10 - \frac {5}{4} \\ & = \frac {35}{4}, \end {align*}
and therefore \(\sum _{n = 3}^{\infty } \frac {1}{n^3} > \frac {7}{96}\), and \(\sum _{n = 1}^{\infty } \frac {1}{n^3} > 1 + \frac {1}{8} + \frac {7}{96} = \frac {115}{96}\).
Hence, \[ \frac {115}{96} < \sum _{n = 1}^{\infty } \frac {1}{n^3} < \frac {116}{96} \] as desired.