Show that the sequence converges to a limit L , the value of which should be stated.

Find the least value of the integer N such that \(\left| {{u_n} - L} \right| < \varepsilon \) , for all n > N where

(i)     \(\varepsilon  = 0.1\);

(ii)     \(\varepsilon  = 0.00001\).

For each of the sequences \(\left\{ {\frac{{{u_n}}}{n}} \right\},{\text{ }}\left\{ {\frac{1}{{2{u_n} - 2}}} \right\}\) and \(\left\{ {{{( - 1)}^n}{u_n}} \right\}\) , determine whether or not it converges.

Prove that the series \(\sum\limits_{n = 1}^\infty  {({u_n} - L)} \) diverges.

\({u_n} = \frac{{3 + \frac{2}{n}}}{{2 - \frac{1}{n}}}\) or \(\frac{3}{2} + \frac{A}{{2n - 1}}\)     M1

using \(\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0\)     (M1)

obtain \(\mathop {\lim }\limits_{n \to \infty } {u_n} = \frac{3}{2} = L\)     A1     N1

[3 marks]

\({u_n} - L = \frac{7}{{2(2n - 1)}}\)     (A1)

\(\left| {{u_n} - L} \right| < \varepsilon  \Rightarrow n > \frac{1}{2}\left( {1 + \frac{7}{{2\varepsilon }}} \right)\)     (M1)

(i)     \(\varepsilon  = 0.1 \Rightarrow N = 18\)     A1

(ii)     \(\varepsilon  = 0.00001 \Rightarrow N = 175000\)     A1

[4 marks]

\({u_n} \to L\) and \(\frac{1}{n} \to 0\)     M1

\( \Rightarrow \frac{{{u_n}}}{n} \to (L \times 0) = 0\) , hence converges     A1

\(2{u_n} - 2 \to 2L - 2 = 1 \Rightarrow \frac{1}{{2{u_n} - 2}} \to 1\) , hence converges     M1A1 

Note: To award A1 the value of the limit and a statement of convergence must be clearly seen for each sequence.

\({( - 1)^n}{u_n}\) does not converge     A1

The sequence alternates (or equivalent wording) between values close to \( \pm L\)     R1

[6 marks]

\({u_n} - L > \frac{7}{{4n}}\) (re: harmonic sequence)     M1

\( \Rightarrow \sum\limits_{n = 1}^\infty  {({u_n} - L)} \) diverges by the comparison theorem     R1 

Note: Accept alternative methods.

[2 marks]

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks. 

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.