The complete supplied teaching material, including deeper learning and source qualifications. Reflection questions here are not automatically graded.
Mathematics A — Week 1
The Mathematics of Logic, Sequences & Series
HSG · Autumn Semester 2026 · Lecture 1 · 15 September
Guided theory: 12–15 minutes · Knowledge check: 5 minutes · Optional deep dives
Three questions connect almost everything in this lesson:
- Logic: When does a mathematical claim actually follow from another?
- Sequences: How does a quantity change over time?
- Series: What happens when we add all those changing quantities together?
By the end, you should be able to interpret implications correctly, understand an induction proof, recognise and analyse sequences, determine when they converge, and calculate finite or infinite geometric sums.
These foundations lead directly into Week 2's financial mathematics.
01 — Mathematical logic: can you trust the conclusion?
Imagine a company makes this promise:
“If a customer spends at least CHF 100, they receive free delivery.”
Define two propositions:
- \(A\): The customer spends at least CHF 100.
- \(B\): The customer receives free delivery.
The promise is written:
\[
A\Rightarrow B
\]
Read it as “If \(A\), then \(B\).”
Here is the subtle part: a customer spends CHF 80 but still receives free delivery. Has the company broken its promise?
No. The company promised what would happen when \(A\) was true. It said nothing about what must happen when \(A\) was false.
The five logical operations
A proposition is a statement that is either true or false.
We build compound propositions using five operations:
A truth table tests every possible combination. Here is the complete comparison:
Two rules matter particularly:
Rule 1: Mathematical OR is inclusive.
\(A\lor B\) remains true when both propositions are true. If exactly one must be true, we need exclusive OR:
\[
(A\lor B)\land\neg(A\land B)
\]
Rule 2: An implication is false in exactly one situation.
\[
\boxed{A\Rightarrow B\text{ is false only if }A=T,\ B=F.}
\]
A useful equivalent expression is:
\[
A\Rightarrow B\quad\equiv\quad\neg A\lor B
\]
Here \(\equiv\) means that the two compound expressions always have the same truth value.
Necessary versus sufficient
If \(A\Rightarrow B\):
- \(A\) is sufficient for \(B\).
- \(B\) is necessary for \(A\).
For example, \(x>5\Rightarrow x>0\). Being greater than 5 is sufficient for being positive. Being positive is necessary for being greater than 5, but it is not sufficient.
The reverse implication does not automatically hold.
\[
A\Rightarrow B\quad\not\Rightarrow\quad B\Rightarrow A
\]
If both directions hold, we have an equivalence:
\[
A\Leftrightarrow B
\quad\Longleftrightarrow\quad
(A\Rightarrow B)\land(B\Rightarrow A)
\]
When a single example destroys a claim
Consider the statement:
“Every growing company is profitable.”
One growing company that loses money is enough to disprove it.
Mathematically, to disprove \(A\Rightarrow B\), find a counterexample in which \(A\) is true and \(B\) is false.
This is an important exam technique: a universal claim can fail because of one valid counterexample.
02 — Mathematical induction: proving infinitely many cases
Checking a formula for \(n=1,2,3\) does not prove that it works for every integer.
Mathematical induction solves this problem with the logic of falling dominoes.
<box border radius="lg" padding={3} gap={2}>
THE DOMINO PRINCIPLE
{#each [1,2,3,4,5] as n}
<box flex="1" align="center" gap={1}>
<box background={n===1?"rgba(16,185,129,0.12)":"surface-secondary"} border radius="sm" padding={{x:2,y:3}} width="100%" align="center">
\(P_{\!{n}}\)
</box>
{#if n===1}<caption>Start</caption>{:else if n===5}<caption>And on…</caption>{:else}<caption>Step</caption>{/if}
</box>
{#if n!==5}{/if}
{/each}
<caption>If the first domino falls, and each falling domino knocks over the next, the entire chain falls.</caption>
</box>
A proof by induction has two essential parts:
1. Base case: Show that \(P(1)\) is true.
2. Inductive step: Assume \(P(n)\) is true for an arbitrary \(n\geq1\), then prove \(P(n+1)\).
Worked example — Sum of the first \(n\) integers
We want to prove:
\[
1+2+\cdots+n=\frac{n(n+1)}2
\]
Base case: \(n=1\).
\[
1=\frac{1(1+1)}2=1
\]
True.
Inductive step: Assume the formula works for \(n\):
\[
1+2+\cdots+n=\frac{n(n+1)}2
\]
Add the next integer to both sides:
\[
\begin{aligned}
1+2+\cdots+n+(n+1)
&=\frac{n(n+1)}2+(n+1)\\
&=\frac{(n+1)(n+2)}2
\end{aligned}
\]
That is exactly the original formula with \(n\) replaced by \(n+1\).
Since the base case and inductive step both hold, the formula is true for all positive integers.
Exam trap: The inductive step must work for every admissible \(n\). A proof that silently relies on overlapping groups, division by zero, or an unstated assumption can be invalid even when the algebra looks convincing.
03 — Sequences: describing change over time
A sequence is an ordered list of numbers:
\[
a_1,a_2,a_3,\ldots
\]
The order matters. The \(n\)-th number is called \(a_n\), and \(n\) is the index.
There are two useful ways to describe a sequence.
Both descriptions generate the same sequence.
An arithmetic sequence adds a constant difference each time. A geometric sequence multiplies by a constant ratio each time.
Geometric sequences are especially useful in economics because percentage growth and decay work multiplicatively.
Worked example — Your investment grows by 5%
You invest CHF 10,000 at time \(t=0\). Each year, the balance grows by 5%.
<box border radius="lg" padding={3} gap={2}>
COMPOUND GROWTH TIMELINE
<grid columns={4} gap={1}>
<grid-item>
<box align="center" gap={1}>
<caption>Year 0</caption>
**10,000**
CHF
</box>
</grid-item>
<grid-item>
<box align="center" gap={1}>
<caption>Year 1</caption>
**10,500**
× 1.05
</box>
</grid-item>
<grid-item>
<box align="center" gap={1}>
<caption>Year 2</caption>
**11,025**
× 1.05
</box>
</grid-item>
<grid-item>
<box align="center" gap={1}>
<caption>Year 20</caption>
**26,533**
CHF, approx.
</box>
</grid-item>
</grid>
<svg viewBox="0 0 340 46" width="100%" xmlns="http://www.w3.org/2000/svg">
<line x1="24" y1="22" x2="316" y2="22" stroke="currentColor" strokeOpacity=".28" strokeWidth="2"/>
<path d="M310 16 L319 22 L310 28" fill="none" stroke="currentColor" strokeWidth="2"/>
{#each [24,112,200,304] as x}
<circle cx={x} cy="22" r="5" fill="#16a34a"/>
{/each}
×1.05
×1.05
···
</svg>
</box>
The recursive rule is:
\[
a_{t+1}=1.05a_t,\qquad a_0=10\,000
\]
Repeated substitution gives:
\[
a_t=10\,000(1.05)^t
\]
After 20 years:
\[
a_{20}=10\,000(1.05)^{20}\approx26\,532.98
\]
Notice the difference between the growth rate and growth factor:
- A 5% increase means multiply by \(1.05\).
- A 5% decrease means multiply by \(0.95\).
This distinction will become central in financial mathematics.
The general geometric sequence
For first term \(a_1\neq0\) and a constant nonzero ratio \(q\):
\[
\frac{a_{n+1}}{a_n}=q
\]
Thus:
\[
\boxed{a_n=a_1q^{n-1}}
\]
Formula status: Derived from the recursive definition in Lecture 1, rather than separately printed as an explicit-term formula on the supplied formulary.
If your sequence starts at \(n=0\), the equivalent expression is \(a_n=a_0q^n\).
The exponent counts the number of multiplication steps, not automatically the calendar year or term number.
04 — Will the sequence settle down?
A sequence can increase, decrease, oscillate, approach a fixed value, or grow without bound.
Three properties help classify its behaviour.
Monotonicity: A sequence is increasing if \(a_{n+1}\geq a_n\) for all \(n\), decreasing if \(a_{n+1}\leq a_n\) for all \(n\), and monotone if either holds throughout.
Boundedness: A sequence is bounded if its terms stay within some fixed interval. In particular, there must be a finite \(M\) such that \(|a_n|\leq M\) for every \(n\).
Convergence: A sequence converges to \(L\) if its terms become arbitrarily close to \(L\) and remain close as \(n\) becomes sufficiently large.
\[
\lim_{n\to\infty}a_n=L
\]
What geometric sequences do
Assume \(a_1>0\). The behaviour depends on the ratio \(q\).
For negative \(a_1\), the increasing/decreasing directions reverse when \(q>0\), but the convergence conclusions remain unchanged.
<box border radius="lg" padding={3} gap={2}>
TWO PATHS: SHRINKING VERSUS GROWING OSCILLATION
<svg viewBox="0 0 340 156" width="100%" xmlns="http://www.w3.org/2000/svg">
{#each [38,111] as y}
<line x1="25" x2="327" y1={y} y2={y} stroke="currentColor" strokeOpacity=".3"/>
{/each}
q = −0.5
q = −1.5
<polyline points="39,15 91,49.5 143,32.3 195,40.9 247,36.6 299,38.7" fill="none" stroke="#16a34a" strokeWidth="2.2" strokeLinejoin="round"/>
<polyline points="39,97 91,132 143,89 195,145 247,76 299,153" fill="none" stroke="#ef4444" strokeWidth="2.2" strokeLinejoin="round"/>
{#each [[39,15],[91,49.5],[143,32.3],[195,40.9],[247,36.6],[299,38.7]] as p}
<circle cx={p[0]} cy={p[1]} r="3.3" fill="#16a34a"/>
{/each}
{#each [[39,97],[91,132],[143,89],[195,145],[247,76],[299,153]] as p}
<circle cx={p[0]} cy={p[1]} r="3.3" fill="#ef4444"/>
{/each}
0
0
</svg>
<caption>Illustrative term plots, independently scaled. Negative ratios flip the sign at every step; the magnitude shrinks for \(|q|<1\) and grows for \(|q|>1\).</caption>
</box>
Two important convergence theorems
Theorem A — Every convergent sequence is bounded.
\[
\text{Convergent}\Rightarrow\text{Bounded}
\]
But the converse is false: \(a_n=(-1)^n\) is bounded and divergent.
Theorem B — Every bounded monotone sequence converges.
\[
\boxed{\text{Bounded + Monotone}\Rightarrow\text{Convergent}}
\]
A sequence does not need to be monotone to converge. For example:
\[
a_n=\frac{(-1)^n}{n}\longrightarrow0
\]
Its terms change sign, but their magnitudes shrink to zero.
The useful distinction is that boundedness and monotonicity together guarantee convergence, while neither condition alone does.
Worked example — Apply the theorem
Consider:
\[
a_n=1-\frac1n,\qquad n\geq1
\]
The first terms are \(0,\frac12,\frac23,\frac34,\ldots\).
Since:
\[
a_{n+1}-a_n=\frac1n-\frac1{n+1}
=\frac1{n(n+1)}>0
\]
the sequence is strictly increasing.
Also \(0\leq a_n<1\), so it is bounded.
Theorem B therefore guarantees convergence. Its limit is:
\[
\lim_{n\to\infty}\left(1-\frac1n\right)=1
\]
The theorem establishes that a limit exists; the final calculation identifies its value.
05 — Series: when you add a sequence together
Suppose an online platform acquires 100 customers in its first month. The number of new customers in each following month is 20% lower than in the previous month.
The monthly new-customer counts form a sequence:
\[
100,\ 80,\ 64,\ 51.2,\ldots
\]
But if you want the total number of customers acquired, you must add the terms.
That produces partial sums:
\[
s_n=\sum_{k=1}^{n}a_k
=a_1+a_2+\cdots+a_n
\]
The sequence of partial sums is called a series in the course's terminology.
Terms versus accumulated totals
<caption>These are expected or modelled quantities; actual customer counts must be whole numbers.</caption>
Notice something important: new customers per month approach zero, while the accumulated total approaches a positive number.
A geometric sequence and its geometric series are different mathematical objects.
The finite geometric-series formula
For \(a_k=a_1q^{k-1}\), with \(q\neq1\):
\[
\boxed{
s_n=a_1\frac{1-q^n}{1-q}
}
\]
Printed on the supplied HSG exam formulary.
For the first four months:
\[
s_4=100\frac{1-0.8^4}{1-0.8}
=295.2
\]
You could add the four numbers individually, but the formula works equally well for 4, 40, or 400 terms.
If \(q=1\), the displayed fraction is undefined. Every term is instead equal to \(a_1\), so:
\[
s_n=na_1
\]
The infinite geometric series
What happens as \(n\to\infty\)?
When \(|q|<1\), we have \(q^n\to0\). Consequently:
\[
\boxed{
\sum_{k=1}^{\infty}a_1q^{k-1}
=\frac{a_1}{1-q}
\qquad(|q|<1)
}
\]
Printed on the supplied HSG exam formulary.
Returning to the platform:
\[
s_\infty=\frac{100}{1-0.8}=500
\]
The model approaches 500 total customer acquisitions, even though it continues indefinitely.
This does not mean that every infinite sum has a finite answer. A nonzero geometric series converges precisely when \(|q|<1\).
For \(q=1.05\), the terms grow rather than shrink, so its infinite series has no finite sum.
Your exam decision sequence
When a problem describes values over time, ask:
- Single value or accumulated total? Use a sequence for one period's value, a series for a sum over periods.
- Where does indexing start? At \(0\), \(1\), or some later date?
- What is the multiplier \(q\)? Translate percentage growth or decay correctly.
- How many terms are included? Count both endpoints when appropriate.
- Is the sum finite or infinite? The infinite formula requires \(|q|<1\).
06 — Your formula-sheet survival guide
Not everything from Week 1 is printed on the supplied exam formulary.
“Printed” means present in the supplied Mathematics A — Formulary, Canvas file 4159731. A formula's appearance in the lecture slides does not necessarily mean it is on the exam formulary.
Five mistakes that cost unnecessary marks
Mistake 1 — Reversing an implication. From \(A\Rightarrow B\), you cannot infer \(B\Rightarrow A\).
Mistake 2 — Confusing bounded with convergent. A bounded sequence can oscillate forever.
Mistake 3 — Ignoring the sign of \(q\). Negative ratios alternate; for example, \(q=-1\) is bounded but divergent.
Mistake 4 — Confusing \(a_n\) and \(s_n\). A single month's revenue is not the same as accumulated revenue.
Mistake 5 — Miscounting the terms. From \(t=0\) through \(t=5\), there are six terms, not five.
Optional deeper understanding
These extensions are useful for proof-based questions. They are not required to follow the main introduction.
A. Why the geometric sum formula works
Let:
\[
s_n=a_1+a_1q+\cdots+a_1q^{n-1}
\]
Multiply by \(q\):
\[
qs_n=a_1q+a_1q^2+\cdots+a_1q^n
\]
Subtract the equations:
\[
s_n-qs_n=a_1-a_1q^n
\]
All the intermediate terms cancel. Hence:
\[
(1-q)s_n=a_1(1-q^n)
\]
For \(q\neq1\):
\[
s_n=a_1\frac{1-q^n}{1-q}
\]
This telescoping argument is more useful than memorising the result without understanding it.
B. What convergence formally means
A sequence converges to \(L\) when, for every tolerance \(\varepsilon>0\), there is an index \(N\) such that:
\[
n\geq N\Rightarrow |a_n-L|<\varepsilon
\]
In other words, however narrow the band around \(L\), every sufficiently late term stays inside it.
For \(a_n=1/n\), the condition \(1/n<\varepsilon\) is satisfied whenever \(n>1/\varepsilon\). Thus \(a_n\to0\).
Notice the phrase every sufficiently late term. Getting close once is not enough.
C. Why an induction proof can fail
In Lecture 1, the professor presents the deliberately false argument that all horses have the same colour.
Its flaw occurs in the transition from one horse to two: the two one-horse groups have no horse in common. The proof implicitly assumes an overlap that does not exist at that step.
The lesson is general: verify the base case and the logic of every inductive transition, rather than accepting a plausible-looking argument.
Check your understanding
Try these five questions before opening the answer key. The objective is to recognise the correct mathematical structure and defend your reasoning.
Question 1 — The business promise
A retailer promises: “If a customer orders at least CHF 150, they receive a gift.”
A customer spends CHF 120 and nevertheless receives a gift.
Has the promise been violated? Does the original promise imply that everyone who receives a gift spent at least CHF 150? Explain both answers using implications.
Question 2 — Build an induction proof
A student claims:
\[
1+3+5+\cdots+(2n-1)=n^2
\]
They verify it for \(n=1,2,3\) and say this proves the formula for all \(n\).
Explain why their argument is incomplete. Then supply the missing general proof using induction.
Question 3 — Oscillation versus convergence
Consider two sequences beginning with \(a_1=8\):
\[
a_{n+1}=-\frac12a_n
\qquad\text{and}\qquad
b_{n+1}=-b_n,\quad b_1=8
\]
Are they bounded, monotone, and convergent? If convergent, determine the limit. Explain why the two sequences behave differently despite both having negative ratios.
Question 4 — The indexing trap
A new service acquires 80 customers in Month 0. The number of new customers doubles each month through Month 3.
Find the number acquired in Month 3 and the cumulative number acquired from Month 0 through Month 3. Explain why using only three terms would be incorrect.
Question 5 — Reverse the model
A company plans three annual bonus payments. The first payment is \(x\), the second is 75% of the first, and the third is 75% of the second.
The total budget for the three payments is CHF 370.
Determine \(x\). Which geometric-series formula applies, and why would the infinite-series formula give the wrong answer?
Explained answer key
Answer 1 — The implication has not been violated
Let \(A\) mean spending at least CHF 150 and \(B\) mean receiving a gift.
The retailer promises \(A\Rightarrow B\).
For this customer, \(A\) is false and \(B\) is true. The implication is therefore true, not violated.
The reverse statement \(B\Rightarrow A\) does not follow. Indeed, this customer is a counterexample: they receive a gift without spending CHF 150.
Key insight: A sufficient condition does not automatically become a necessary condition.
Answer 2 — Three checked cases do not prove an infinite claim
The student checked only three values. To prove the claim for every positive integer, we need the base case and a general inductive step.
For \(n=1\):
\[
1=1^2
\]
Assume:
\[
1+3+\cdots+(2n-1)=n^2
\]
The next odd number is \(2n+1\). Adding it gives:
\[
n^2+(2n+1)=(n+1)^2
\]
Therefore, whenever the claim holds for \(n\), it also holds for \(n+1\).
Together with the base case, this proves the formula for all \(n\geq1\).
Key insight: Induction proves a general transition, not merely a collection of examples.
Answer 3 — Both oscillate, but only one settles down
The first sequence has \(q=-1/2\) and terms:
\[
8,-4,2,-1,\frac12,\ldots
\]
It is bounded, not monotone, and converges to zero because \(|q|<1\).
The second sequence has \(q=-1\) and terms:
\[
8,-8,8,-8,\ldots
\]
It is bounded and not monotone, but does not converge. The distance from zero never shrinks, and the terms continue alternating between two distinct values.
Key insight: Alternating signs do not prevent convergence. Persistent non-shrinking oscillation does.
Answer 4 — There are four months
The monthly counts are:
Thus Month 3 has 640 new customers.
The cumulative total is:
\[
80+160+320+640=1\,200
\]
Equivalently, using the finite geometric sum:
\[
s_4=80\frac{1-2^4}{1-2}=1\,200
\]
There are four terms because the indexed months are \(0,1,2,3\).
Key insight: The final index is not necessarily the number of terms.
Answer 5 — The first payment is CHF 160
The three payments are:
\[
x,\quad 0.75x,\quad 0.75^2x
\]
Their sum must equal 370:
\[
x(1+0.75+0.75^2)=370
\]
Since \(1+0.75+0.75^2=2.3125\):
\[
x=\frac{370}{2.3125}=160
\]
Check: \(160+120+90=370\).
The finite geometric-sum formula applies because exactly three payments are made.
Using \(x/(1-0.75)\) would incorrectly include infinitely many future payments.
Key insight: A decreasing payment stream is not automatically a perpetuity. The number of payments still matters.
Ready for Week 1 exercises
You now have the foundations needed to approach the official Exercise Series 1.
The HSG HS2026 course schedule specifically identifies these common-class priorities: A2 — truth tables; A6 — applications of geometric sequences and series; A11 — limits of sequences.
Your next step is to apply these ideas to unfamiliar questions, not simply repeat the worked examples.
Sources and coverage
- Primary: Enrico De Giorgi, Mathematics with Statistics A — Lecture 1: Mathematical Logic, Sequences, and Series, 15 September 2026. Verified PDF sections: mathematical logic (pp. 17–41), sequences (pp. 43–64), series (pp. 66–74), and summary (pp. 79–81). Official English lecture PDF.
- Formulary: Mathematics A — Formulary, Canvas file 4159731, section Analysis: Geometric Sequences and Series. Both finite and infinite geometric-sum formulas are printed there.
- Schedule: Mathematics with Statistics A — How is the Lecture Organized?, HS2026 English course-information PDF, pp. 9–10, for Week 1 topic and Exercise Series 1 mapping.
- Scope: The examples, questions, conceptual diagrams, and answers on this page are original Monastery teaching material. They are not represented as official HSG exam questions. The core Week 1 coverage is supported by the identified official lecture and schedule; complete coverage of all Canvas folders and later-released materials has not been independently verified.