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]