Foundations — Geometric series — convergence condition, proof

This page unpacks every symbol and idea the parent note Geometric series leans on, starting from things a smart 12-year-old already knows. Nothing here assumes you've met "series", "sigma", or "limit" before — we build each from a picture.

1. The first term $a$, and multiplying by a fixed factor

Before "series", the simplest object is a list of numbers where each one comes from the one before by multiplying by the same number every time. To describe such a list we only need two things: where it starts and what it multiplies by.

Look at the figure. The leftmost bar is the first term $a$; each following bar is $r$ times the height of the bar to its left. When $r$ is between $0$ and $1$, the bars shrink — that shrinking is the whole story of convergence.

2. Powers and the notation $r^n$

Writing $r\cdot r\cdot r$ gets tiring, so we use an exponent.

The one that trips people up: $==r^0=1==$. Why? Zero copies of $r$ multiplied together is the "empty product", and multiplying nothing leaves you at the multiplication-starting-point, which is $1$. This matters because a series that starts at $n=0$ has first term $a\cdot r^0=a$.

3. Absolute value $|r|$ — the size of a number

The convergence rule is stated with $|r|$, not $r$. We need this because a negative $r$ can still make the pieces shrink.

The figure lays $r$ on a horizontal number line (the axis running left–right through zero). The blue double-arrow measures the distance from $0$ out to $-3$, and the orange double-arrow measures the distance from $0$ out to $+3$. Both arrows have the same length, $3$ — which is exactly the picture of $|-3|=|3|=3$: the arrows point in opposite directions, but only their length is what $|r|$ reports.

4. The summation symbol $\sum$

Now the piece the parent note uses everywhere.

Written out in full, that same sum is ${\sum }_{n=0}^{N-1}a{r}^n=\underset{n=0}{\underbrace{a{r}^0}}+\underset{n=1}{\underbrace{a{r}^1}}+\underset{n=2}{\underbrace{a{r}^2}}+\cdots +\underset{n=N-1}{\underbrace{a{r}^{N-1}}}.$

So $\sum _{n=0}^{2}a{r}^n=a{r}^0+a{r}^1+a{r}^2=a+ar+a{r}^2$ (small enough that no dots are needed). It's just shorthand for a long "$+++$".

5. Partial sum $S_N$ — stopping early on purpose

We can never add infinitely many numbers by hand. The trick is to add only the first few, then watch a pattern.

The figure shows the running total climbing as we add more terms. For shrinking pieces (blue) the running total flattens toward a ceiling; for growing pieces (orange) it shoots up forever. The parent note's whole proof is about finding a clean formula for this climbing total and then asking where it ends up.

6. The limit $\underset{N\to \infty }{\lim }$ — "where is it heading?"

This is the one genuinely new idea, and the entire convergence question rides on it.

The single fact the proof needs is the behaviour of $r^N$ as $N\to \infty$:

• $|r|<1$: each multiply by $r$ makes it smaller in size, so $r^N\to 0$. (Pieces vanish.)
• $|r|>1$: each multiply makes it bigger, so $|r^N|\to \infty$. (No destination.)
• $r=1$: $r^N=1$ forever. (Never shrinks.)
• $r=-1$: $r^N$ flips $1,-1,1,-1,\dots$. (Bounces, never settles.)

Every case in the figure. Notice: only when the dots march into zero (the $|r|<1$ curves) can the total settle down. This is exactly why the convergence condition is $|r|<1$.

7. Two small algebra tools the proof reuses

Factoring. Pulling a common piece out front: $ax-ay=a(x-y)$. The proof uses $a-a{r}^N=a(1-r^N)$ and $S_N-r{S}_N=S_N(1-r)$.

The term-by-term cancellation (WHY the middle vanishes). The proof's key move is to compute $S_N-r{S}_N$. Write the two sums stacked, aligning matching powers of $r$:

Now subtract the bottom row from the top row, column by column:

• The $a$ in $S_N$ has nothing beneath it in $r{S}_N$ → it survives.
• The $ar$ appears in both rows → $ar-ar=0$, gone.
• The $a{r}^2$ appears in both rows → $a{r}^2-a{r}^2=0$, gone.
• Every middle term $a{r}^k$ (for $k=1,2,\dots ,N-1$) shows up once in each row, because multiplying $S_N$ by $r$ just shifts every term up one power — so each middle term meets an identical twin and cancels.
• The $a{r}^N$ in $r{S}_N$ has nothing above it in $S_N$ → it survives, carrying its minus sign.

So the entire middle collapses and only the two ends are left: $S_N-r{S}_N=a-a{r}^N.$

Dividing to isolate. Factor both sides: $S_N(1-r)=a(1-r^N)$. Dividing both sides by $(1-r)$ frees $S_N$: $S_N=\frac{a(1-r^N)}{1-r}.$ This division is only legal when $1-r\ne 0$, i.e. $r\ne 1$ — which is exactly why the parent note treats $r=1$ separately.

8. Closing the loop — what the infinite series actually sums to

Everything above exists to answer one question: as $N\to \infty$, where does $S_N$ land? Feed the limit of $r^N$ into the partial-sum formula.

Prerequisite map

The diagram below traces how each foundation feeds the next, ending at the topic's two headline results. Read it top-to-bottom: the fixed multiplier and its powers give us notation ($\sum$) and a way to measure shrinkage ($|r|$); those build the partial sum; the partial sum plus the limit deliver the convergence condition and the sum formula.

Each box on the left must be solid before the topic's proof (the two bottom boxes) makes sense.

Equipment checklist

Cover the right side and answer aloud — if any stumps you, reread that section before the parent note.

Connections

• Sequences — limits and convergence — where $\lim r^N$ and "settling on a number" come from.
• Partial sums and series convergence — the idea that a series is the limit of its partial sums.
• Ratio Test — generalises the "constant ratio" to ratios that change.
• Power series and radius of convergence — geometric series is the model case $\sum x^n=\frac{1}{1-x}$.
• Repeating decimals as fractions — a direct payoff once these foundations are in place.
• Telescoping series — the cancellation trick that turns $S_N-r{S}_N$ into a clean formula.