Series Expansion for Pi Cosecant of Pi Lambda

From ProofWiki
Jump to navigation Jump to search

Theorem

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 $x = 0$:

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


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

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

and thus:

\(\ds \) \(\) \(\ds \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} } + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n - \lambda} + \frac 1 {n - 1 - \lambda} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n + \lambda} + \frac 1 {n - 1 - \lambda} }\)

The result follows.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Series_Expansion_for_Pi_Cosecant_of_Pi_Lambda&oldid=540810"