Engineering Mathematics: Geometric Series

$a + ar + a r^2 + a r^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r}$

Proof

$s=a + ar + a r^2 + a r^3 + \cdots + a r^{n-1}$

Multiply s with r , therefore

$rs=ar + ar^2 + a r^3 + a r^4 + \cdots + a r^{n}$

Subtract rs from s , therefore

$s-rs = s(1-r) = a-ar^n$

therefore,

$s= a\frac{1-r^{n}}{1-r}$

As n goes to infinity, the absolute value of r must be less than one for the series to converge, therefore ,

$s= a\frac{1}{1-r}$

$S= e^{-p}+e^{-2p}+e^{-3p}+e^{-4p}+\cdots+e^{-np} = \sum_{k=1}^ne^{-kp}$

As n goes to infinity, the absolute value of e converges

$S=\frac{e^{-p}}{1-e^{-p}}=\frac{1}{e^p-1}$

$\sum _{k=0}^{\infty }k^{2}q^{k-1}=\frac{2q}{(1-q)^{3}}+\frac{1}{(1-q)^{2}}$

Given

$\sum _{k=0}^{\infty }q^{k}=\frac{1}{1-q}$

(1)

First derivative

$\sum _{k=0}^{\infty }kq^{k-1}=\frac{1}{(1-q)^{2}}$

(2)

Second derivative

$\sum _{k=0}^{\infty }k(k-1)q^{k-2}=\frac{2}{(1-q)^{3}}$

(3)

$\sum _{k=0}^{\infty }k(k-1)q^{k-2}=\sum _{k=0}^{\infty }k^{2}q^{k-2}-\sum _{k=0}^{\infty }kq^{k-2}$

(4)

$\sum _{k=0}^{\infty }k(k-1)q^{k-2}=\frac{1}{q}\sum _{k=0}^{\infty }k^{2}q^{k-1}-\frac{1}{q}\sum _{k=0}^{\infty }kq^{k-1}$

(5)

from (3) and (5)

$\frac{2}{(1-q)^{3}}=\frac{1}{q}\sum _{k=0}^{\infty }k^{2}q^{k-1}-\frac{1}{q}\sum _{k=0}^{\infty }kq^{k-1}$

(6)

from (2) and (6)

$\frac{2}{(1-q)^{3}}=\frac{1}{q}\sum _{k=0}^{\infty }k^{2}q^{k-1}-\frac{1}{q}\frac{1}{(1-q)^{2}}$

(7)

$\frac{1}{q}\sum _{k=0}^{\infty }k^{2}q^{k-1}=\frac{2}{(1-q)^{3}}+\frac{1}{q}\frac{1}{(1-q)^{2}}$

(8)

$\sum _{k=0}^{\infty }k^{2}q^{k-1}=\frac{2q}{(1-q)^{3}}+\frac{1}{(1-q)^{2}}$

Engineering Mathematics