2022 DHS Promo Q10

By considering \(4r-1=5r-\left( 1+r \right)\), use the method of differences to find \(\sum\limits_{r=1}^{n}{\left( \frac{4r-1}{{{5}^{r}}} \right)}\).[4]

Hence find \(\sum\limits_{r=1}^{\infty }{\left( \frac{4r-1}{{{5}^{r}}} \right)}\), justifying your answer.[2]

Given that \(\sum\limits_{r=1}^{\infty }{{{a}_{r}}}\) and \(\sum\limits_{r=1}^{\infty }{{{b}_{r}}}\) are series with non-negative terms, such that \({{a}_{r}}<{{b}_{r}}\) for all \(r\), the Comparison Test states that if \(\sum\limits_{r=1}^{\infty }{{{b}_{r}}}\) converges then \(\sum\limits_{r=1}^{\infty }{{{a}_{r}}}\) converges.

By considering your result in part (b), use the Comparison Test to show that \(\sum\limits_{r=1}^{\infty }{\left( \frac{r-1}{{{6}^{r}}} \right)}\) converges.[2]