Engineering Mathematics: Apr 4, 2014

DI.ID.3: Reciprocal series and digamma function relationship

$\sum _{n=1}^{\infty }\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{b-a}$

Given

$\Psi (z)= -\gamma -\int _{0}^{1}\frac{t^{z}-1}{1-t}$

(1)

$\frac{1}{(n+a)(n+b)} = \frac{A}{(n+a)} - \frac{B}{(n+b)}$

(2)

$\frac{A}{(n+a)}=\int _{0}^{\infty }e^{-\frac{(n+a)x}{A}}dx$

$\frac{B}{(n+b)}=\int _{0}^{\infty }e^{-\frac{(n+b)x}{B}}dx$

(3)

$\newline A(n+a)-B(n+b)=1 \newline A*n=B*n \newline A*b-B*b=1 \newline A=B \newline A(b-a)=1 \newline A= \frac{1}{b-a}\newline$

(4)

Proof

$\frac{A}{(n+a)}-\frac{B}{(n+b)}=\int _{0}^{\infty }e^{-\frac{(n+a)x}{A}}dx-\int _{0}^{\infty }e^{-\frac{(n+b)x}{B}}dx$

$\frac{A}{(n+a)}-\frac{B}{(n+b)}=\int _{0}^{\infty }e^{-\frac{nx}{A}}e^{-\frac{ax}{A}}dx-\int _{0}^{\infty }e^{-\frac{nx}{B}}e^{-\frac{bx}{B}}dx$

(5)

$\sum _{n=1}^{\infty }\left ( \frac{A}{(n+a)}-\frac{B}{(n+b)} \right )=\sum _{n=1}^{\infty }\left ( \int _{0}^{\infty }e^{-\frac{nx}{A}}e^{-\frac{ax}{A}}dx-\int _{0}^{\infty }e^{-\frac{nx}{B}}e^{-\frac{bx}{B}}dx \right )$

(6)

$\sum _{n=1}^{\infty }\left ( \frac{A}{(n+a)}-\frac{B}{(n+b)} \right )=\left ( \int _{0}^{\infty }\left [ \frac{e^{-\frac{x}{A}}}{1-e^{-\frac{x}{A}}} \right ]e^{-\frac{ax}{A}}dx-\int _{0}^{\infty }\left [ \frac{e^{-\frac{x}{B}}}{1-e^{-\frac{x}{B}}} \right ]e^{-\frac{bx}{B}}dx \right )$

(7)

$\newline e^{-\frac{x}{A}} =t\newline e^{-\frac{x}{A}}=t^{a}\newline \frac{-1}{A}e^{-\frac{x}{A}}dx=dt \rightarrow dx= -\frac{A}{t}\: dt$

(8)

$\sum _{n=1}^{\infty }\left ( \frac{A}{(n+a)}-\frac{B}{(n+b)} \right )=\left ( \int _{0}^{1 }\left [ \frac{t}{1-t} \right ]t^{a}\frac{A}{t}\: dt-\int _{0}^{1 }\left [ \frac{t}{1-t} \right ]t^{b}\frac{B}{t}\: dt \right )$