Given that \(n > {\text{ln}}\,n\) for \(n > 0\), use the comparison test to show that the series \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) is divergent.

Find the interval of convergence for \(\sum\limits_{n = 0}^\infty  {\frac{{{{\left( {3x} \right)}^n}}}{{{\text{ln}}\left( {n + 2} \right)}}} \).

METHOD 1

\({\text{ln}}\left( {n + 2} \right) < n + 2\)     (A1)

\( \Rightarrow \frac{1}{{{\text{ln}}\left( {n + 2} \right)}} > \frac{1}{{n + 2}}\) (for \(n \geqslant 0\))     A1

Note: Award A0 for statements such as \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}}  > \sum\limits_{n = 0}^\infty  {\frac{1}{{n + 2}}} \). However condone such a statement if the above A1 has already been awarded.

\(\sum\limits_{n = 0}^\infty  {\frac{1}{{n + 2}}} \) (is a harmonic series which) diverges     R1

Note: The R1 is independent of the A1s.

Award R0 for statements such as "\(\frac{1}{{n + 2}}\) diverges".

so \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) diverges by the comparison test     AG

METHOD 2

\(\frac{1}{{{\text{ln}}\,n}} > \frac{1}{n}\) (for \(n \geqslant 2\))     A1

Note: Award A0 for statements such as \(\sum\limits_{n = 2}^\infty  {\frac{1}{{{\text{ln}}\,n}}}  > \sum\limits_{n = 2}^\infty  {\frac{1}{n}} \). However condone such a statement if the above A1 has already been awarded.

a correct statement linking \(n\) and \(n + 2\) eg,

\(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}}  = \sum\limits_{n = 2}^\infty  {\frac{1}{{{\text{ln}}\,n}}} \) or \(\sum\limits_{n = 0}^\infty  {\frac{1}{{n + 2}}}  = \sum\limits_{n = 2}^\infty  {\frac{1}{n}} \)     A1

Note: Award A0 for \(\sum\limits_{n = 0}^\infty  {\frac{1}{n}} \)

\(\sum\limits_{n = 2}^\infty  {\frac{1}{n}} \) (is a harmonic series which) diverges

(which implies that \(\sum\limits_{n = 2}^\infty  {\frac{1}{{{\text{ln}}\,n}}} \) diverges by the comparison test)      R1

Note: The R1 is independent of the A1s.

Award R0 for statements such as \(\sum\limits_{n = 0}^\infty  {\frac{1}{n}} \) deiverges and "\({\frac{1}{n}}\) diverges".

Award A1A0R1 for arguments based on \(\sum\limits_{n = 1}^\infty  {\frac{1}{n}} \).

so \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) diverges by the comparison test      AG

[3 marks]

applying the ratio test \(\mathop {{\text{lim}}}\limits_{n \to \infty } \left| {\frac{{{{\left( {3x} \right)}^{n + 1}}}}{{{\text{ln}}\left( {n + 3} \right)}} \times \frac{{{\text{ln}}\left( {n + 2} \right)}}{{{{\left( {3x} \right)}^n}}}} \right|\)     M1

\( = \left| {3x} \right|\) (as \(\mathop {{\text{lim}}}\limits_{n \to \infty } \left| {\frac{{{\text{ln}}\left( {n + 2} \right)}}{{{\text{ln}}\left( {n + 3} \right)}}} \right| = 1\)     A1

Note: Condone the absence of limits and modulus signs.

Note: Award M1A0 for \(3{x^n}\). Subsequent marks can be awarded.

series converges for \( - \frac{1}{3} < x < \frac{1}{3}\)

considering \(x =  - \frac{1}{3}\) and \(x = \frac{1}{3}\)     M1

Note: Award M1 to candidates who consider one endpoint.

when \(x = \frac{1}{3}\), series is \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) which is divergent (from (a))      A1

Note: Award this A1 if \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) is not stated but reference to part (a) is.

when \(x =  - \frac{1}{3}\), series is \(\sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}}}{{{\text{ln}}\left( {n + 2} \right)}}} \)     A1

\(\sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}}}{{{\text{ln}}\left( {n + 2} \right)}}} \) converges (conditionally) by the alternating series test      R1

(strictly alternating, \(\left| {{u_n}} \right| > \left| {{u_{n + 1}}} \right|\) for \(n \geqslant 0\) and \(\mathop {{\text{lim}}}\limits_{n \to \infty } \left( {{u_n}} \right) = 0\))

so the interval of convergence of S is \( - \frac{1}{3} \leqslant x < \frac{1}{3}\)     A1

Note: The final A1 is dependent on previous A1s – ie, considering correct series when \(x =  - \frac{1}{3}\) and \(x = \frac{1}{3}\) and on the final R1.

Award as above to candidates who firstly consider \(x =  - \frac{1}{3}\) and then state conditional convergence implies divergence at \(x = \frac{1}{3}\).

[7 marks]