# Sequences and series: exercise sheet 7

Exercise.

Prove that if $a n + 1 = k 1 + a n$ where $k > 0$ and $a 1 > 0$, then the sequence $( a n )$ converges to the positive root of $x 2 + x = k$.

NOTE: The word discuss is used to frame deliberately open-ended questions. Say as much as you reasonably can in the time available. All your assertions should be proved. Discuss the convergence of these sequences clearly is asking for an indication of whether the sequences do or do not converge, and if possible to what.

Solution to Exercise 2(a) For $n ∈ ℕ$, $1 + 1 n > 1 + 1 n + 1$ hence $( 1 + 1 n ) 1 2 > ( 1 + 1 n + 1 ) 1 2$ so the sequence is monotonic decreasing. Each value $( 1 + 1 n ) 1 2$ is greater than zero because $1 + 1 n > 0$ and the exponent $1 2$ means take the positive square root, so the sequence is bounded below by $0$. Therefore by the Monotone Convergence Theorem, $a n → l$ for some $l ⩾ 0$ in $ℝ$. By continuity of the square root function and $1 + 1 n → 1$, $l = ( lim 1 + 1 n ) 1 2 = 1$. (Note: in this case the monotone convergence theorem was not actually needed as the limit can be found and proved correct by continuity. But in other cases such as 2(e) or 3(e) where the limit is not so easy to evaluate you may like to use Monotone Convergence.)