Sequences & Series

10 methods · 34 questions

★ Signature method unlocked: the technique this chapter rewards most often, with 3 worked questions.

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When asked to sum a finite or infinite series whose term has no obvious closed form, try to write that term as $g (r) - g (r + 1)$ (or $g (r) - g (r + k)$ ): the sum then collapses to the surviving endpoints $g (1) - g (N + 1)$ , and the infinite sum is $g (1) - lim g (N + 1)$ .

Trigger: a sum to evaluate exactly, especially a product in the denominator, a $tan / arctan$ term, or a difference you are handed in a 'by considering ...' hint.

Instances:
(i) partial fractions on a rational term, e.g. $\frac{1}{r ( r + 1 ) ( r + 2 )}$ or $\frac{2}{r ( r ^{2} - 1 )}$ , split so consecutive pieces cancel;
(ii) a trigonometric term turned into a difference by a product-to-sum identity, e.g. $2 sin \frac{1}{2} x cos r x = sin (r + \frac{1}{2}) x - sin (r - \frac{1}{2}) x$ , so $\sum cos r x$ telescopes;
(iii) an $arctan$ term written as $arctan u_{r} - arctan u_{r + 1}$ using $arctan a - arctan b = arctan \frac{a - b}{1 + ab}$ ;
(iv) a factored difference handed to you, $\frac{1}{1 + x ^{r}} - \frac{1}{1 + x ^{r + 1}}$ , whose terms you recognise as the summand.

Linked questions (3)

STEP 3 2012 · Q2

In this question,

∣ x ∣ < 1

and you may ignore issues of convergence.

Simplify $(1 - x) (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}}),$ where $n$ is a positive integer, and deduce that $\frac{1}{1 - x} = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}}) + \frac{x ^{2^{n + 1}}}{1 - x} .$ Deduce further that $ln (1 - x) = - r = 0 \sum \infty ln (1 + x^{2^{r}}),$ and hence that $\frac{1}{1 - x} = \frac{1}{1 + x} + \frac{2 x}{1 + x ^{2}} + \frac{4 x ^{3}}{1 + x ^{4}} + \dots .$
Show that $\frac{1 + 2 x}{1 + x + x ^{2}} = \frac{1 - 2 x}{1 - x + x ^{2}} + \frac{2 x - 4 x ^{3}}{1 - x ^{2} + x ^{4}} + \frac{4 x ^{3} - 8 x ^{7}}{1 - x ^{4} + x ^{8}} + \dots .$

STEP 3 2012 · Q8

The sequence

F_{0}

F_{1}

F_{2}

\dots

is defined by

F_{0} = 0

F_{1} = 1

and, for

n ⩾ 0

F_{n + 2} = F_{n + 1} + F_{n} .

Show that $F_{0} F_{3} - F_{1} F_{2} = F_{2} F_{5} - F_{3} F_{4}$ .
Find the values of $F_{n} F_{n + 3} - F_{n + 1} F_{n + 2}$ in the two cases that arise.
Prove that, for $r = 1$ , $2$ , $3$ , $\dots$ , $arctan (\frac{1}{F _{2 r}}) = arctan (\frac{1}{F _{2 r + 1}}) + arctan (\frac{1}{F _{2 r + 2}})$ and hence evaluate the following sum (which you may assume converges): $r = 1 \sum \infty arctan (\frac{1}{F _{2 r + 1}}) .$

STEP 2 2014 · Q6

By simplifying

sin (r + \frac{1}{2}) x - sin (r - \frac{1}{2}) x

or otherwise show that, for

sin \frac{1}{2} x \neq = 0

cos x + cos 2 x + \dots + cos n x = \frac{sin ( n + \frac{1}{2} ) x - sin \frac{1}{2} x}{2 sin \frac{1}{2} x} .

The functions

S_{n}

, for

n = 1

2

, …, are defined by

S_{n} (x) = r = 1 \sum n \frac{1}{r} sin r x (0 ⩽ x ⩽ π) .

Find the stationary points of $S_{2} (x)$ for $0 ⩽ x ⩽ π$ , and sketch this function.
Show that if $S_{n} (x)$ has a stationary point at $x = x_{0}$ , where $0 < x_{0} < π$ , then $sin n x_{0} = (1 - cos n x_{0}) tan \frac{1}{2} x_{0}$ and hence that $S_{n} (x_{0}) ⩾ S_{n - 1} (x_{0})$ . Deduce that if $S_{n - 1} (x) > 0$ for all $x$ in the interval $0 < x < π$ , then $S_{n} (x) > 0$ for all $x$ in this interval.
Prove that $S_{n} (x) ⩾ 0$ for $n ⩾ 1$ and $0 ⩽ x ⩽ π$ .