\(% 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 the first identity, notice that \begin {align*} \frac {1}{r + 1} - \frac {1}{r} + \frac {1}{r^2} & = \frac {r^2 - r(r + 1) + (r + 1)}{r^2 (r + 1)} \\ & = \frac {r^2 - r^2 - r + r + 1}{r^2 (r + 1)} \\ & = \frac {1}{r^2 (r + 1)}, \end {align*}
and hence using this, \begin {align*} \sum _{r = 1}^{N} \frac {1}{r^2 (r + 1)} & = \sum _{r = 1}^{N} \left (\frac {1}{r + 1} - \frac {1}{r} + \frac {1}{r^2}\right ) \\ & = \sum _{r = 1}^{N} \frac {1}{r^2} + \sum _{r = 1}^{N} \frac {1}{r + 1} - \sum _{r = 1}^{N} \frac {1}{r} \\ & = \sum _{r = 1}^{N} \frac {1}{r^2} + \sum _{r = 2}^{N + 1} \frac {1}{r} - \sum _{r = 1}^{N} \frac {1}{r} \\ & = \sum _{r = 1}^{N} \frac {1}{r^2} - \frac {1}{1} + \frac {1}{N + 1} \\ & = \sum _{r = 1}^{N} \frac {1}{r^2} - 1 + \frac {1}{N + 1}. \end {align*}
Let \(N \to \infty \), and we have \(\frac {1}{N + 1} \to 0\), and hence \[ \sum _{r = 1}^{\infty } \frac {1}{r^2 (r + 1)} = \sum _{r = 1}^{\infty } \frac {1}{r^2} - 1 = \frac {\pi ^2}{6} - 1. \]
By partial fractions, let \[ \frac {1}{r^2 (r + 1) (r + 2)} = \frac {Ar + B}{r^2} + \frac {C}{r + 1} + \frac {D}{r + 2} \] for real constants \(A, B, C\) and \(D\).
Hence, \[ (Ar + B) (r + 1) (r + 2) + C r^2 (r + 2) + D r^2 (r + 1) = 1. \]
Let \(r = -2\), we have \(D \cdot (-2)^2 \cdot (-1) = -4D = 1\), and hence \(D = - \frac {1}{4}\).
Let \(r = -1\), we have \(C \cdot (-1)^2 \cdot 1 = C = 1\), and hence \(C = 1\).
Let \(r = 0\), we have \(B \cdot 1 \cdot 2 = 1\), and hence \(B = \frac {1}{2}\).
Considering the coefficient of \(r^3\), we have \(A + C + D = 0\), and hence \(A = - \frac {3}{4}\).
Hence, \[ \frac {1}{r^2 (r + 1) (r + 2)} = - \frac {3}{4} \cdot \frac {1}{r} + \frac {1}{2} \cdot \frac {1}{r^2} + \frac {1}{r + 1} - \frac {1}{4} \cdot \frac {1}{r + 2}. \]
Therefore, \begin {align*} \sum _{r = 1}^{N} \frac {1}{r^2 (r + 1) (r + 2)} & = - \frac {3}{4} \sum _{r = 1}^{N} \frac {1}{r} + \frac {1}{2} \sum _{r = 1}^{N} \frac {1}{r^2} + \sum _{r = 1}^{N} \frac {1}{r + 1} - \frac {1}{4} \sum _{r = 1}^{N} \frac {1}{r + 2} \\ & = \frac {1}{2} S_{N} - \frac {3}{4} \cdot \sum _{r = 1}^{N} \frac {1}{r} + \sum _{r = 2}^{N + 1} \frac {1}{r} - \frac {1}{4} \sum _{r = 3}^{N + 2} \frac {1}{r} \\ & = \frac {1}{2} S_{N} - \frac {3}{4} \sum _{r = 3}^{N} \frac {1}{r} + \sum _{r = 3}^{N} \frac {1}{r} - \frac {1}{4} \sum _{r = 3}^{N} \frac {1}{r} \\ & = \frac {1}{2} S_{N} - \frac {3}{4} \left (\frac {1}{1} + \frac {1}{2}\right ) + \left (\frac {1}{2} + \frac {1}{N + 1}\right ) - \frac {1}{4} \left (\frac {1}{N + 1} + \frac {1}{N + 2}\right ) \\ & = \frac {1}{2} S_{N} - \frac {9}{8} + \frac {4}{8} + \frac {3}{4} \cdot \frac {1}{N + 1} - \frac {1}{4} \cdot \frac {1}{N + 2} \\ & = \frac {1}{2} S_{N} - \frac {5}{8} + \frac {3}{4} \cdot \frac {1}{N + 1} - \frac {1}{4} \cdot \frac {1}{N + 2}. \end {align*}
Let \(N \to \infty \), we have \(\frac {1}{N + 1}, \frac {1}{N + 2} \to 0\), and hence \[ \sum _{r = 1}^{\infty } \frac {1}{r^2 (r + 1) (r + 2)} = \frac {1}{2} \lim _{N \to \infty } S_{N} - \frac {5}{8} = \frac {\pi ^2}{12} - \frac {5}{8}. \]
Similarly, let \[ \frac {1}{r^2 (r + 1)^2} = \frac {A}{r^2} + \frac {B}{r} + \frac {C}{(r + 1)^2} + \frac {D}{r + 1} \] for some real constants \(A, B, C\) and \(D\).
Hence, \[ 1 = A (r + 1)^2 + B r (r + 1)^2 + C r^2 + D r^2 (r + 1). \]
Let \(r = 0\), and we have \(A = 1\). Let \(r = -1\), and we have \(C = 1\). Considering the coefficient of \(r^3\) we have \(B + D = 0\), and for \(r\), \(2A + B = 0\).
Hence, \(B = -2, D = 2\), and \[ \frac {1}{r^2 (r + 1)^2} = \frac {1}{r^2} - \frac {2}{r} + \frac {1}{(r + 1)^2} + \frac {2}{r + 1}. \]
Therefore, for natural numbers \(N\), we have \begin {align*} \sum _{r = 1}^{N} \frac {1}{r^2 (r + 1)^2} & = \sum _{r = 1}^{N} \frac {1}{r^2} + \sum _{r = 1}^{N} \frac {1}{(r + 1)^2} + 2 \sum _{r = 1}^{N} \frac {1}{r + 1} - 2 \sum _{r = 1}^{N} \frac {1}{r} \\ & = S_N + \sum _{r = 1}^{N + 1} \frac {1}{r^2} + 2 \sum _{r = 2}^{N + 1} \frac {1}{r} - 2 \sum _{r = 1}^{N} \frac {1}{r} \\ & = S_{N} + S_{N + 1} - \frac {1}{1^2} + 2 \cdot \frac {1}{N + 1} - 2 \cdot 1 \\ & = S_{N} + s_{N + 1} + 2 \cdot \frac {1}{N + 1} - 3. \end {align*}
Let \(N \to \infty \). \(S_{N}, S_{N + 1} \to \frac {\pi ^2}{6}\), and \(\frac {1}{N + 1} \to 0\). Hence, \begin {align*} \sum _{r = 1}^{\infty } \frac {1}{r^2 (r + 1)^2} & = 2 \cdot \frac {\pi ^2}{6} - 3 \\ & = \frac {\pi ^2}{3} - 3 \\ & = 2 \cdot \left (\frac {\pi ^2}{6} - 1\right ) - 1 \\ & = 2 \sum _{r = 1}^{\infty } \frac {1}{r^2 (r + 1)} - 1 \\ & = \sum _{r = 1}^{\infty } \frac {2}{r^2 (r + 1)} - 1, \end {align*}
as desired.