\(% 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}}\)
Notice that \[ \frac {1}{1 + x^r} - \frac {1}{1 + x^{r + 1}} = \frac {x^{r + 1} - x^r}{(1 + x^r)(1 + x^{r + 1})} = \frac {x^r (x - 1)}{(1 + x^r)(1 + x^{r + 1})}. \]
Therefore, we have \begin {align*} \sum _{r = 1}^{N} \frac {x^r}{(1 + x^r)(1 + x^{r + 1})} & = \sum _{r = 1}^{N} \frac {1}{x - 1} \left [\frac {1}{1 + x^r} - \frac {1}{1 + x^{r + 1}}\right ] \\ & = \frac {1}{x - 1} \sum _{r = 1}^{N}\left [\frac {1}{1 + x^r} - \frac {1}{1 + x^{r + 1}}\right ] \\ & = \frac {1}{x - 1} \left [\frac {1}{1 + x} - \frac {1}{1 + x^{n + 1}}\right ]. \end {align*}
For \(\abs *{x} < 1\), as \(n \to \infty \), \(x^{n + 1} \to 0\). Therefore, \begin {align*} \sum _{r = 1}^{\infty } \frac {x^r}{(1 + x^r)(1 + x^{r + 1})} & = \frac {1}{x - 1} \left [\frac {1}{1 + x} - 1\right ] \\ & = \frac {1}{x - 1} \cdot \frac {-x}{1 + x} \\ & = \frac {x}{1 - x^2} \end {align*}
as desired.
Notice that \begin {align*} \sech (ry) \sech ((r + 1)y) & = \frac {2}{e^{ry} + e^{-ry}} \cdot \frac {2}{e^{(r + 1)y} + e^{-(r + 1)y}} \\ & = \frac {4e^{-ry - (r + 1)y}}{\left (1 + e^{-2ry}\right ) \left (1 + e^{-2(r+1)y}\right )} \\ & = 4 e^{-y} \frac {e^{-2ry}}{\left (1 + e^{-2ry}\right ) \left (1 + e^{-2(r+1)y}\right )}. \end {align*}
Let \(x = e^{-2y}\). We have \[ \sech (ry) \sech ((r + 1)y) = 4e^{-y} \frac {x^r}{\left (1 + x^r\right )\left (1 + x^{r + 1}\right )}. \]
When \(y > 0\), \(x = e^{-2y} \in (0, 1)\). Therefore, \begin {align*} \sum _{r = 1}^{\infty } \sech (ry) \sech ((r + 1)y) & = 4 e^{-y} \frac {e^{-2y}}{1 - e^{-4y}} \\ & = 2e^{-y} \frac {2}{e^{2y} - e^{-2y}} \\ & = 2e^{-y} \cosech (2y) \end {align*}
as desired.
Notice that for all \(x \in \RR \), \(\cosh x = \cosh (-x)\), therefore \(\sech x = \sech (-x)\).
Therefore, \begin {align*} & \phantom {=} \sum _{r = -\infty }^{\infty } \sech (ry) \sech ((r + 1)y) \\ & = \sum _{r = 1}^{\infty } \sech (ry) \sech ((r + 1)y) + \sum _{r = -\infty }^{0} \sech (ry) \sech ((r + 1)y) \\ & = \sum _{r = 1}^{\infty } \sech (ry) \sech ((r + 1)y) + \sum _{r = 0}^{\infty } \sech (-ry) \sech ((-r+1)y) \\ & = \sum _{r = 1}^{\infty } \sech (ry) \sech ((r + 1)y) + \sum _{r = 0}^{\infty } \sech (ry) \sech ((r-1)y) \\ & = \sum _{r = 1}^{\infty } \sech (ry) \sech ((r + 1)y) + \sum _{r = 2}^{\infty } \sech (ry) \sech ((r-1)y) + \sech (y) \sech (0) + \sech (0) \sech (-y) \\ & = \sum _{r = 1}^{\infty } \sech (ry) \sech ((r + 1)y) + \sum _{r = 1}^{\infty } \sech ((r + 1)y) \sech (ry) + 2\sech y \\ & = 4e^{-y} \cosech (2y) + 2 \sech y \\ & = \frac {4e^{-y}}{\sinh 2y} + \frac {2}{\cosh y} \\ & = \frac {2e^{-y}}{\sinh y \cosh y} + \frac {2}{\cosh y} \\ & = \frac {2e^{-y} + 2 \sinh y}{\sinh y \cosh y} \\ & = \frac {2 e^{-y} + e^{y} - e^{-y}} {\sinh y \cosh y} \\ & = \frac {e^y - e^{-y}}{\sinh y \cosh y} \\ & = \frac {2 \cosh y}{\sinh y \cosh y} \\ & = 2 \cosech y. \end {align*}