Prove by induction that \(n! > {3^n}\), for \(n \ge 7,{\text{ }}n \in \mathbb{Z}\).

Hence use the comparison test to prove that the series \(\sum\limits_{r = 1}^\infty  {\frac{{{2^r}}}{{r!}}} \) converges.

if \(n = 7\) then \(7! > {3^7}\)     A1

so true for \(n = 7\)

assume true for \(n = k\)     M1

so \(k! > {3^k}\)

consider \(n = k + 1\)

\((k + 1)! = (k + 1)k!\)     M1

\( > (k + 1){3^k}\)

\( > 3.3k\;\;\;({\text{as }}k > 6)\)     A1

\( = {3^{k + 1}}\)

hence if true for \(n = k\) then also true for \(n = k + 1\). As true for \(n = 7\), so true for all \(n \ge 7\).     R1

Note:     Do not award the R1 if the two M marks have not been awarded.

[5 marks]

consider the series \(\sum\limits_{r = 7}^\infty  {{a_r}} \), where \({a_r} = \frac{{{2^r}}}{{r!}}\)     R1

Note:     Award the R1 for starting at \(r = 7\)

compare to the series \(\sum\limits_{r=7}^\infty  {{b_r}} \) where \({b_r} = \frac{{{2^r}}}{{{3^r}}}\)     M1

\(\sum\limits_{r = 7}^\infty  {{b_r}} \) is an infinite Geometric Series with \(r = \frac{2}{3}\) and hence converges     A1

Note:     Award the A1 even if series starts at \(r = 1\).

as \(r! > {3^r}\) so \((0 < ){a_r} < {b_r}\) for all \(r \ge 7\)     M1R1

as \(\sum\limits_{r = 7}^\infty  {{b_r}} \) converges and \({a_r} < {b_r}\) so \(\sum\limits_{r = 7}^\infty  {{a_r}} \) must converge

Note:     Award the A1 even if series starts at \(r = 1\).

as \(\sum\limits_{r = 1}^6 {{a_r}} \) is finite, so \(\sum\limits_{r = 1}^\infty  {{a_r}} \) must converge     R1

Note:     If the limit comparison test is used award marks to a maximum of R1M1A1M0A0R1.

[6 marks]

Total [11 marks]