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]