Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman
Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
$\displaystyle \displaystyle\huge\frac{x}{(1-x)\left(1-x^2\right)}+\frac{x^2}{\left(1-x^2\right)\left(1-x^3\right)}+\frac{x^3}{\left(1-x^3\right)\left(1-x^4\right)}+.....\frac{x^k}{\left(1-x^k\right)\left(1-x^{k+1}\right)}$
$\displaystyle =\displaystyle\huge\frac{1}{1-x}\left[\frac{x}{\left(1-x^2\right)}+\frac{x^2(1-x)}{\left(1-x^2\right)\left(1-x^3\right)}+\frac{x^3(1-x)}{\left(1-x^3\right)\left(1-x^4\right)}+....\frac{x^k(1-x)}{\left(1-x^k\right)\left(1-x^{k+1}\right)}\right]$
$\displaystyle =\displaystyle\huge\frac{1}{1-x}\left[\frac{x}{\left(1-x^2\right)}+\frac{x^2-x^3}{\left(1-x^2\right)\left(1-x^3\right)}+\frac{x^3-x^4}{\left(1-x^3\right)\left(1-x^4\right)}+...\right]$
$\displaystyle =\displaystyle\huge\frac{1}{1-x}\left[\frac{x}{1-x^2}+\frac{x^2}{1-x^2}-\frac{x^3}{1-x^3}+\frac{x^3}{1-x^3}-\frac{x^4}{1-x^4}+\frac{x^4}{1-x^4}-.....\right]$
$\displaystyle =\displaystyle\huge\frac{1}{1-x}\left[\frac{x(1+x)}{(1-x)(1+x)}-\frac{x^{k+1}}{1-x^{k+1}}\right]=\frac{x}{(1-x)^2}-\frac{x^{k+1}}{(1-x)\left(1-x^{k+1}\right)}$
We have $\displaystyle \frac{1}{1-x^k}-\frac{1}{1-x^{k+1}} = \frac{x^k-x^{k+1}}{(1-x^k)\cdot (1-x^{k+1})}= (1-x)\cdot \frac{x^k}{(1-x^k)\cdot (1-x^{k+1})}$.
Hence choosing $\displaystyle f(k)=\tfrac{1}{1-x}\cdot \frac{1}{1-x^k}$ works because $\displaystyle f(k)-f(k+1)=\frac{x^k}{(1-x^k)\cdot (1-x^{k+1})}$
Then $\displaystyle \sum_{k=1}^n\frac{x^k}{(1-x^k)\cdot (1-x^{k+1})} = \sum_{k=1}^n\left(f(k)-f(k+1)\right)=f(1)-f(n+1)=\tfrac{1}{1-x}\cdot \left(\tfrac{1}{1-x}-\tfrac{1}{1-x^{n+1}}\right)$