# Series Expansion for Pi Cosecant of Pi Lambda

## Theorem

Let $\lambda \in \R \setminus \Z$ be a real number which is not an integer.

Then:

$\displaystyle \pi \csc \pi \lambda = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n + \lambda} + \frac 1 {n - 1 - \lambda} }$

## 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:

$\displaystyle \cos \lambda x \sim \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\lambda^2 - n^2} }$

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 $x = 0$:

 $\displaystyle \cos 0$ $=$ $\displaystyle \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos 0} {\lambda^2 - n^2} }$ $\displaystyle \leadsto \ \$ $\displaystyle \frac \pi {\sin \lambda \pi}$ $=$ $\displaystyle 2 \lambda \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac 1 {\lambda^2 - n^2} }$ Cosine of Zero is One and rearrangement $\displaystyle \leadsto \ \$ $\displaystyle \pi \csc \lambda \pi$ $=$ $\displaystyle \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \lambda} {\lambda^2 - n^2}$ Definition of Real Cosecant Function $\displaystyle$ $=$ $\displaystyle \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\lambda + n + \lambda - n} {\paren {\lambda + n} \paren {\lambda - n} }$ Difference of Two Squares $\displaystyle$ $=$ $\displaystyle \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\lambda - n} {\paren {\lambda + n} \paren {\lambda - n} } + \frac {\lambda + n} {\paren {\lambda + n} \paren {\lambda - n} } }$ $\displaystyle$ $=$ $\displaystyle \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} }$

This expression is fine as is, but to obtain the form we set out to prove, we observe:

 $\displaystyle$  $\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} + \frac 1 {n - 1 - \lambda} }$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} } - \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \paren {\frac 1 {n - 1 - \lambda} }$ Both converge by Power Series Expansion for Logarithm of 1 + x $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} } - \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} }$ Translation of Index Variable of Summation $\displaystyle$ $=$ $\displaystyle - \paren {-1}^0 \paren {\frac 1 {0 - \lambda} }$ $\displaystyle$ $=$ $\displaystyle \frac 1 \lambda$

and thus:

 $\displaystyle$  $\displaystyle \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} }$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} } + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} + \frac 1 {n - 1 - \lambda} }$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n + \lambda} + \frac 1 {n - 1 - \lambda} }$

The result follows.

$\blacksquare$