Given that \(\mathrm{f}\left( r \right)=\frac{r}{2{{\,}^{r}}}\), by considering \(\mathrm{f}\left( r+1 \right)-\mathrm{f}\left( r \right)\), find \(\sum\limits_{r=\,1}^{n}{\frac{1-r}{2{{\,}^{r+1}}}}\).[3]
Cauchy’s root test states that a series of the form \(\sum\limits_{r=0}^{\infty }{{{a}_{r}}}\) (where \({{a}_{r}}>0\) for all \(r\)) converges when \(\mathop{\lim}\limits_{n\to \infty }\,\sqrt[n]{{{a}_{n}}}<1\), and diverges when \(\mathop{\lim}\limits_{n\to \infty }\,\sqrt[n]{{{a}_{n}}}>1\). When \(\mathop{\lim}\limits_{n\to \infty }\,\sqrt[n]{{{a}_{n}}}=1\), the test is inconclusive. Using the test and given that \(\mathop{\lim}\limits_{n\to \infty }\,\sqrt[n]{{{n}^{p}}}=1\) for all positive \(p\), explain why the series \(\sum\limits_{r=0}^{\infty }{\frac{2{{\,}^{r}}{{r}^{x}}}{{{3}^{\,r}}}}\) converges for all positive values of \(x\).[3]
By considering \({{\left( 1-y \right)}^{-2}}=1+2y+3{{y}^{2}}+4{{y}^{3}}+...\), evaluate \(\sum\limits_{r=0}^{\infty }{\frac{2{{\,}^{r}}{{r}^{x}}}{{{3}^{\,r}}}}\) for the case when \(x=1\).[2]