Find the interval of convergence of the series $\sum\limits_{k = 1}^\infty  {\frac{{{{(x - 3)}^k}}}{{{k^2}}}}$.

ratio test $\mathop {\lim }\limits_{k \to \infty } \left| {\frac{{{u_{k + 1}}}}{{{u_k}}}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{{{(x - 3)}^{k + 1}}{k^2}}}{{{{(k + 1)}^2}{{(x - 3)}^k}}}} \right|$     M1A1

$\mathop {\lim }\limits_{k \to \infty } \left| {(x - 3)\frac{{{k^2}}}{{{{(k + 1)}^2}}}} \right|$

Note: Condone absence of limits and modulus signs in above.

$\left| {x - 3} \right|\mathop {\lim }\limits_{k \to \infty } \left| {{{\left( {\frac{1}{{1 + \frac{1}{k}}}} \right)}^2}} \right| - \left| {x - 3} \right|$     A1

for convergence $\left| {x - 3} \right| < 1$     (M1)

$\Rightarrow  - 1 < x - 3 < 1$

$\Rightarrow 2 < x < 4$     (A1)

now we need to test end points.     (M1)

when $x = 4$ we have $\sum\limits_{k = 1}^\infty  {\frac{1}{{{k^2}}}}$ which is a convergent series     R1

when $x = 2$ we have $\sum\limits_{k = 1}^\infty  {\frac{{{{( - 1)}^k}}}{{{k^2}}} =  - 1 + \frac{1}{{{2^2}}} - \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} +  \ldots }$ which is convergent     R1

(alternating series/absolutely converging series)

hence the interval of convergence is $[2,{\text{ }}4]$     A1

This question was well answered in general although the presentation was sometimes poor.