Using l’Hôpital’s rule, find $\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\arcsin \left( {\frac{1}{{\sqrt {(x + 1)} }}} \right)}}{{\frac{1}{{\sqrt x }}}}} \right)$.

(i) Find ${a_1}$ and ${a_2}$ and hence write down an expression for ${a_n}$.

(ii) Show that ${\theta _n} = \arcsin \frac{1}{{\sqrt {(n + 1)} }}$.

Using a suitable test, determine whether this series converges or diverges.

$\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\arcsin \left( {\frac{1}{{\sqrt {(x + 1)} }}} \right)}}{{\frac{1}{{\sqrt x }}}}} \right)$ is of the form $\frac{0}{0}$

and so will equal the limit of $\frac{{\frac{{\frac{{ - 1}}{2}{{(x + 1)}^{ - \frac{3}{2}}}}}{{\sqrt {1 - \left( {\frac{1}{{x + 1}}} \right)} }}}}{{\frac{{ - 1}}{2}{x^{ - \frac{3}{2}}}}}$     M1M1A1A1

Note: M1 for attempting differentiation of the top and bottom, M1A1 for derivative of top (only award M1 if chain rule is used), A1 for derivative of bottom.

$= \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\frac{x}{{(x + 1)}}} \right)}^{\frac{3}{2}}}}}{{\sqrt {\frac{x}{{x + 1}}} }} = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{x}{{x + 1}}} \right)$    M1

Note: Accept any intermediate tidying up of correct derivative for the method mark.

$= 1$    A1

[6 marks]

(i)     ${a_1} = \sqrt 2 ,{\text{ }}{a_2} = \sqrt 3$     A1

${a_n} = \sqrt {n + 1}$    A1

(ii)     $\sin {\theta _n} = \frac{1}{{{a_n}}} = \frac{1}{{\sqrt {n + 1} }}$     A1

Note: Allow ${\theta _n} = \arcsin \left( {\frac{1}{{{a_n}}}} \right)$ if ${a_n} = \sqrt {n + 1}$ in b(i).

so ${\theta _n} = \arcsin \frac{1}{{\sqrt {(n + 1)} }}$     AG

[3 marks]

for $\sum\limits_{n = 1}^\infty  {\arcsin \frac{1}{{\sqrt {(n + 1)} }}}$ apply the limit comparison test (since both series of positive terms)     M1

with $\sum\limits_{n = 1}^\infty  {\frac{1}{{\sqrt n }}}$     A1

from (a) $\mathop {\lim }\limits_{n \to \infty } \frac{{\arcsin \frac{1}{{\sqrt {(n + 1)} }}}}{{\frac{1}{{\sqrt n }}}} = 1$, so the two series either both converge or both diverge     M1R1

$\sum\limits_{n = 1}^\infty  {\frac{1}{{\sqrt 2 }}}$ diverges (as is a $p$-series with $p = \frac{1}{2}$)     A1

hence $\sum\limits_{n = 1}^\infty  {{\theta _n}}$ diverges     A1

[6 marks]