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]