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.