2022 DHS Promo Q10
(a)
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]
(b)
Hence find $\sum\limits_{r=1}^{\infty }{\left( \frac{4r-1}{{{5}^{r}}} \right)}$, justifying your answer.
[2]
(c)
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]
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