Engineering Mathematics: Cauchy Product (Series convolution )

This post will help students understand the derivation of harmonic series through Cauchy Product

Given identities:

Logarithmic Function

$\ln (1-x)=-\sum _{n=0}^{\infty }\frac{x^{n}}{n}=-(1+x+\frac{x^{2}}{2}+\cdot \cdot \cdot +\frac{x^{n}}{n})$

(1)

Geometric Series

$\frac{1}{1-x}=\sum _{n=0}^{\infty }x^{n}=1+x+x^{2}+\cdot \cdot \cdot +x^{n}$

(2)

Harmonic Series

$H_{n}=\sum _{k=0}^{n}\frac{1}{k}=H_{n}= 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdot \cdot \cdot+\frac{1}{n}$

(3)

product of (1) and (2)

$\frac{-\ln (1-x)}{1-x}=\left ( \sum _{n=1}^{\infty }\frac{x^{n}}{n} \right )\left ( \sum _{n=0}^{\infty }x^{n} \right )$

based on (SC.ID.1)

$a_{n}=\frac{1}{n}$

$b_{n}=1$

therefore,

$c_{n}=\sum _{k=0}^{n}\frac{1}{k}=H_{n}$

therefore,

$\frac{-\ln (1-x)}{1-x}=\sum _{n=0}^{\infty }H_{n}x^{n}$

(4)

if x=1/2

$\sum _{n=0}^{\infty }\frac{H_{n}}{2^{n+1}}=\ln (2)$

Integration of (4)

$-\int _{0}^{x}\frac{\ln (1-x)}{1-x}=\int _{0}^{x}\sum _{n=0}^{\infty }H_{n}x^{n}$

$\ln^{2} (1-x)=2\sum _{n=0}^{\infty }H_{n}\frac{x^{n+1}}{n+1}$

Engineering Mathematics