Use method of difference to find sum of series

[chwala]

Homework Statement

see attached.

Relevant Equations

series sum

My interest is on the (highlighted part in yellow ) of finding the partial fractions- Phew took me time to figure out this out

My approach on the highlighted part;

i let

##(kr+1) =x ##

then, ##\dfrac{1}{(kr+1)(kr-k+1)} = \dfrac{1}{x(x-k)}##
then,

##\dfrac{1}{(x)(x-k)}=\dfrac{A}{x} +\dfrac{B}{x-k}##

##1=A(x-k)+Bx##

##Ax+Bx=0##
##-Ak=1##

##A=\dfrac{-1}{k}##

i know that ##A+B=0##
##B=\dfrac{1}{k}##

thus,

##\dfrac{1}{(x)(x-k)}=\dfrac{-1}{kx} +\dfrac{1}{k(x-k)}##

##\dfrac{1}{(kr+1)(kr-k+1)}=\dfrac{-1}{k(kr+1)} +\dfrac{1}{k(kr-k+1)}##

##\dfrac{1}{(kr+1)(kr-k+1)}= \dfrac{1}{k(kr-k+1)}-\dfrac{1}{k(kr+1)}##

##\dfrac{1}{(kr+1)(kr-k+1)}= \dfrac{1}{k(k(r-1)+1)}-\dfrac{1}{k(kr+1)}##

any simpler or alternative approach (in determining the partial fractions) prompted this post.

 

[chwala]

This is also related,

solution:

My steps,

##\lim_{n \rightarrow +\infty} \left[\dfrac{n}{kn+1}\right] = \lim_{n \rightarrow +\infty} \left[\dfrac{1}{k+1/n}\right]= \left[\dfrac{1}{k+0}\right]=\dfrac{1}{k}##