Theorem: Series Expansion for Pi Cosecant of Pi Lambda

Let $\lambda \in \R \setminus \Z$ be a real number which is not an integer.
Then:
:$\ds \pi \csc \pi \lambda = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n + \lambda} + \frac 1 {n - 1 - \lambda} }$

Proof:
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
:$\map f x = \cos \lambda x$
From Half-Range Fourier Cosine Series: $\cos \lambda x$ over $\openint 0 \pi$ its Fourier series can be expressed as:
:$\ds \cos \lambda x \sim \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\lambda^2 - n^2} }$
Because of the nature of this expansion, we have that:
:$\map f {0^+} = \map f {0^-}$
and so the expansion holds for $x = 0$.
So, setting $x = 0$:
{{begin-eqn}}
{{eqn | l = \cos 0
      | r = \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos 0} {\lambda^2 - n^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac \pi {\sin \lambda \pi}
      | r = 2 \lambda \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac 1 {\lambda^2 - n^2} }
      | c = Cosine of Zero is One and rearrangement
}}
{{eqn | ll= \leadsto
      | l = \pi \csc \lambda \pi
      | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \lambda} {\lambda^2 - n^2}
      | c = {{Defof|Real Cosecant Function}}
}}
{{eqn | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\lambda + n + \lambda - n} {\paren {\lambda + n} \paren {\lambda - n} }
      | c = Difference of Two Squares
}}
{{eqn | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\lambda - n} {\paren {\lambda + n} \paren {\lambda - n} } + \frac {\lambda + n} {\paren {\lambda + n} \paren {\lambda - n} } }
      | c = 
}}
{{eqn | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} }
      | c = 
}}
{{end-eqn}}
This expression is fine as is, but to obtain the form we set out to prove, we observe:
{{begin-eqn}}
{{eqn | o =
      | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} + \frac 1 {n - 1 - \lambda} }
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} } - \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \paren {\frac 1 {n - 1 - \lambda} }
      | c = Both converge by Power Series Expansion for Logarithm of 1 + x
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} } - \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} }
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = - \paren {-1}^0 \paren {\frac 1 {0 - \lambda} }
}}
{{eqn | r = \frac 1 \lambda
}}
{{end-eqn}}
and thus:
{{begin-eqn}}
{{eqn | o =
      | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} }
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} } + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} + \frac 1 {n - 1 - \lambda} }
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n + \lambda} + \frac 1 {n - 1 - \lambda} }
}}
{{end-eqn}}
The result follows.
{{qed}}