Show that the series is conditionally convergent but not absolutely convergent.

Show that \(S > 0.4\) .

\(\sin \left( {\frac{1}{n}} \right)\) decreases as n increases     A1

\(\sin \left( {\frac{1}{n}} \right) \to 0\) as \(n \to \infty \)     A1

so using the alternating series test, the series is conditionally convergent     R1

comparing (the absolute series) with the harmonic series:

\(\sum\limits_{n = 1}^\infty  {\frac{1}{n}} \)     M1

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {\frac{1}{n}} \right)}}{{\frac{1}{n}}} = 1\)     A1

since the harmonic series is divergent, it follows by the limit comparison theorem that the given series is not absolutely convergent     R1

hence the series is conditionally convergent     AG

[6 marks]

successive partial sums are

\(0.841…\)

\(0.362…\)

\(0.689…\)

\(0.441…\)     A1

since \(S\) lies between any pair of successive partial sums, it follows that \(S\) lies between \(0.441…\) and \(0.689…\)     R1

and is therefore greater than \(0.4\)     AG

Note: Use of the facts that the error is always less than the modulus of the next term, or the sequence of even partial sums gives lower bounds are equally acceptable.

[2 marks]