Sequences and Series

MAT syllabus

Sequences defined iteratively and by formulae. Arithmetic and geometric progressions*. Their sums*. Convergence condition for infinite geometric progressions*.

* Part of full A-level Mathematics syllabus.

Revision

A sequence $a_n$ might be defined by a formula for the $n^\text{th}$ term like $a_n=n^2-n$.
A sequence $a_n$ might be defined with an relation like $a_{n+1}=f(a_n)$ for $n\geq 0$, if we're given the function $f(x)$ and also given a first term like $a_0=1$. (The "first term" might instead be called $a_0$ if we feel like counting from zero).
The sum of the first $n$ terms of a sequence $a_k$ can be written with the notation $\displaystyle \sum_{k=0}^{n-1} a_k$ (if the first term is $a_0$) or $\displaystyle \sum_{k=1}^{n} a_k$ (if the first term is $a_1$).
An arithmetic sequence is one where the difference between terms is constant. The terms can be written as $a$ ,$a+d$, $a+2d$, $a+3d$, $\dots$ where $a$ is the first term and $d$ is the common difference.
The sum of the first $n$ terms of an arithmetic sequence with first term $a$ and common difference $d$ is $\displaystyle\frac{n}{2}\left(2a+(n-1)d\right)$, which you can remember as "first term plus last term, times the number of terms, divided by two".
A geometric sequence is one where the ratio between consecutive terms is constant. The terms can be written as $a$, $ar$, $ar^2$, $ar^3$, $\dots$ where $a$ is the first term and $r$ is the common ratio.
The sum of the first $n$ terms of a geometric sequence with first term $a$ and common ratio $r$ is $\displaystyle \frac{a(1-r^n)}{1-r}$. One way to remember this is to remember what happens if we multiply the sum of the first $n$ terms of a geometric series by $(1-r)$,
\begin{align*}
(1-r)(a+ar+\dots +ar^{n-1})=&(a-ar)+(ar-ar^2)+\dots+(ar^{n-1}-ar^n)\\
=&a-ar^n.
\end{align*}
For a geometric sequence $a_n$, the sum to infinity is written as $\displaystyle \sum_{k=0}^\infty a_k$. If the common ratio $r$ satisfies $|r|<1$ then this is equal to $\displaystyle \frac{a}{1-r}$. If $|r|\geq 1$ then this sum to infinity does not converge (it does not approach any particular real number).

Warm-up

A sequence is defined by $a_n=n^2-n$. What is $a_3$? What is $a_{10}$? Find $a_{n+1}-a_n$ in terms of $n$. Find $a_{n+1}-2a_n+a_{n-1}$ in terms of $n$.
A sequence is defined by $a_0=1$ and $a_n=a_{n-1}+3$ for $n\geq 1$. Find $a_0+a_1+\dots + a_{10}$. Find $a_{1000}$.
A sequence is defined by $a_0=1$ and $a_n=\frac{a_{n-1}}{3}$ for $n\geq 1$. Find $a_0+a_1+\dots + a_{10}$. Find $a_{1000}$. Does the sum of all the terms of this sequence converge? If it does, what is the sum to infinity?
A sequence is defined by $a_0=1$ and $a_n=3a_{n-1}+1$ for $n\geq 1$. A sequence $b_n$ is defined by $b_n=A \times 3^n+B$ where $A$ and $B$ are real numbers. Find values for $A$ and $B$ such that $a_n=b_n$ for all $n\geq 0$.
A sequence is defined by $a_n=An^2+Bn+C$ where $A$, $B$, and $C$ are real numbers. Find $A$, $B$, and $C$ in terms of $a_0$, $a_1$, and $a_2$. Hint: you'll need to solve 3 simultaneous equations.
Simplify $2^1+2^2+2^3+\dots +2^n$ for $n\geq 1$.
Simplify $3^4+3^5+3^6+\dots + 3^n$ for $n\geq 4$.
When does the sum $1+x^3+x^6+x^9+x^{12}+...$ converge? Simplify it in the case that it converges.
When does the sum $\displaystyle 2-x+\frac{x^2}{2}-\frac{x^3}{4}+\dots$ converge? Simplify it in the case that it converges.
Consider the sum of the first $n$ terms of an arithmetic sequence $a_1$, $a_2$, $\dots$, $a_n$ with $a_1=a$ and $a_2=a+d$. Explain why the sum of the $i^{\text{th}}$ term and the $(n+1-i)^\text{th}$ term doesn't depend on $i$, as long as $1\leq i \leq n$. By considering separate cases where $n$ is even or where $n$ is odd, deduce that the sum of the first $n$ terms of an arithmetic sequence is $n$ times the average term.
Consider the sum of the first $n$ terms of an arithmetic sequence with first term $a$ and constant difference $d$. Consider the special case $d=0$. Write down the sum in this case. Now consider the case $a=0$. In this case, write the sum in terms of the triangle numbers $T_n=1+2+3+\dots+n=\frac{1}{2} n(n+1)$. Hence write down the sum of the first $n$ terms of an arithmetic sequence. Check that this agrees with the formula above.

MAT questions

MAT 2016 Q1A

A sequence $ a_n$ has first term $a_1 = 1$, and subsequent terms defined by $a_{n+1} = la_n$ for $n \geqslant 1$. What is the product of the first 15 terms of the sequence?

(a) $l^{14}$,

(b) $15 + l^{14}$,

(d) $l^{105}$,

(e) $15 + l^{105}$.

Hint: note that this question is asking for the product and not the sum. Also note that the first term is $a_1$ and not $a_0$, so the first 15 terms will be $a_1$, $a_2$, $\dots$, $a_{14}$, $a_{15}$.

MAT 2016 Q1G

The sequence $x_{n}$, where $n\geqslant 0$, is defined by $x_{0}=1$ and
\[
x_{n}=\sum_{k=0}^{n-1}x_{k}\qquad \text{for }n\geqslant 1.
\]
The sum
\[
\sum_{k=0}^{\infty }\frac{1}{x_{k}}
\]
equals

(a) $1$,

(b) $\displaystyle \frac{6}{5}$,

(d) $3$,

(e) $\displaystyle \frac{27}{5}$.

Hint: work out a few of the values $x_1$, $x_2$, $x_3$, $\dots$ before trying to work out the sum to infinity.

MAT 2017 Q1C

A sequence $(a_n)$ has the property that
\begin{equation*}
a_{n+1} = \frac{a_{n}}{a_{n-1}}
\end{equation*}
for every $n \geqslant 2$. Given that $a_1 = 2$ and $a_2 = 6$, what is $a_{2017}$?

(a) $\displaystyle\frac{1}{6}$,

(b) $\displaystyle\frac{2}{3}$,

(d) $\displaystyle 2$,

(e) $\displaystyle 3$.

Hint: again, work out a few of the values $x_1$, $x_2$, $x_3$, $\dots$.

MAT 2016 Q5

This question concerns the sum $s_n$ defined by
$$s_n = 2 + 8 + 24 + \cdots + n2^n.$$

(i) Let $f(n) = (An+B)2^n + C$ for constants $A$, $B$ and $C$ yet to be determined, and suppose $s_n = f(n)$ for all $n\geq1$. By setting $n=1,2,3$, find equations that must be satisfied by $A$, $B$ and $C$.

(ii) Solve the equations from part (i) to obtain values for $A$, $B$ and $C$.

(iii) Using these values, show that if $s_k = f(k)$ for some $k\geq1$ then $s_{k+1} = f(k+1)$.

You may now assume that $f(n)=s_n$ for all $n\geq1$.

(iv) Find simplified expressions for the following sums:\begin{align*}
t_n&= n + 2(n-1) + 4(n-2) + 8(n-3) + \cdots + 2^{n-1}1,\\
u_n&=\frac12+\frac24 + \frac38 + \cdots + \frac{n}{2^n}.
\end{align*}

(v) Find the sum \begin{equation*}\sum_{k=1}^n s_k.\end{equation*}

Hints: At the start, take a moment to understand the definition of $s_n$. How do the numbers 8 and 24 relate to the $+n2^n$ part of the definition? What are the values of $s_1$ and $s_2$ and $s_3$? (Hint; $s_2$ is not 8).

In part (iii) we're being asked to investigate what happens when we go from $s_{k}$ to $s_{k+1}$. From the definition at the top, what changes when we go from $s_{k}$ to $s_{k+1}$? If we do that to $f(k)$, do we get to $f(k+1)$?

In part (iv), it would be good if we could find a link between $t_n$ and $s_n$, perhaps by spotting a copy of the sum that defines $s_n$ hiding in there. Then we want to find a link between $u_n$ and $s_n$, or a link between $u_n$ and $t_n$. If that doesn't work, we can go back to the idea in part (i) and try to find a general expression for the $n^\text{th}$ term of $t_n$ or $u_n$ by guessing a function like $(An+B)2^n+C$ or maybe like $(An+B)2^{-n}+C$.

In part (v), we know an expression for $s_k$ in terms of things like $2^k$ and $k2^k$. We know how to sum the first of those things, and the sum of the second thing there is oddly familiar from earlier in this question...

Extension

A future session of the Oxford MAT Livestream will be on "recursion", and we'll look at more expressions that are like $a_n=f(a_{n-1})$ but more complex.

The following material is included for your interest only, and not for MAT preparation.

There's a general formula for sequences where the difference between terms is itself an arithmetic sequence. The sequences are sometimes called quadratic sequences, and they have $a_n=An^2+Bn+C$ for some $A$, $B$, and $C$. You can probably guess what happens if the difference between terms of a sequence is itself a quadratic sequence.

In MAT 2016 Q5, we found a formula for the sum of the first $n$ terms of the sequence $a_k= k2^k$ with one particular method (guess the formula, check the formula). Here's a more direct proof; expand and sum and sum.

Expand out each term into $2^k$s, so that we've got\begin{equation*}
\sum_{k=1}^n k2^k=(2^1)+(2^2+2^2)+(2^3+2^3+2^3)+\dots+(\underbrace{2^n+2^n+\dots+2^n}_{n}).
\end{equation*}
Regroup the terms and sum\begin{align*}
\sum_{k=1}^n k2^k=& \left(2^1+2^2+\dots+2^n\right)+(2^2+2^3+\dots+2^n)+\dots+(2^{n-1}+2^n)+2^n\\
=& 2^1(2^{n}-1)+2^2(2^{n-1}-1)+\dots+2^{n-1}(2^2-1)+2^n(2^1-1)
\end{align*}
Expand these brackets, bring together all the $2^{n+1}$s and notice that the remaining terms are another geometric series\begin{align*}
\sum_{k=1}^n k2^k=& n2^{n+1}-2^1-2^2-2^3-\dots -2^n\\
=& n2^{n+1}-2^1(2^n-1)\\
=& (2n-2)2^n+2
\end{align*}

You might be able to adapt this method to similar sums $\sum_{k=1}^n kx^k$ for other numbers $x$. (Watch out for a factor of $(x-1)$ from the geometric sums, which is 1 above when $x=2$.)

For the sum to infinity, if $|x|<1$, then we get $\displaystyle x+2x^2+3x^3+4x^4+\dots=\frac{x}{(1-x)^2}$. This agrees with a different, more advanced calculation using calculus; the sum is\begin{equation*}
x+2x^2+3x^3+4x^4+\dots=x \frac{\mathrm{d}}{\mathrm{d}x}\left(1+x+x^2+x^3+\dots\right)=x\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{1-x}\right) =\frac{x}{(1-x)^2}
\end{equation*}
but this calculation uses the "chain rule" for differentiation, which is not on the MAT syllabus!