2019 SAJC P2 Q4
(a)
Given that $\text{f}\left( r \right)=\frac{r}{2{{\,}^{r}}}$, by considering $\text{f}\left( r+1 \right)-\text{f}\left( r \right)$, find $\sum\limits_{r=\,1}^{n}{\frac{1-r}{2{{\,}^{r+1}}}}$.
[3]
(b)
(i) 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 $\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}}<1$, and diverges when $\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}}>1$. When $\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}}=1$, the test is inconclusive. Using the test and given that $\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{n}^{p}}}=1$ 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]
(ii) 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]
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