The number e — definition as limit of (1+1 - n)ⁿ, natural growth context: Visual walkthrough

Definition What a "limit" means (we use it from Step 4 on)

When we write $lim_{n \to \infty} f (n) = L$ we mean something completely concrete: as you push $n$ to larger and larger whole numbers, the values $f (n)$ get and stay as close to the fixed number $L$ as you please — you can name any tiny gap and, past some point, every value sits inside it. In this walkthrough $f (n) = (1 + \frac{1}{n})^{n}$ and $L$ is the ceiling $e$ . Why we're allowed to work term-by-term. Later we treat each ingredient's limit separately — the messy factor $(1 - \frac{j}{n})$ , the sum, and so on. That is only legal when the piece is a fixed, finite expression (a fixed power, a fixed product, a fixed number of added terms): for those, "limit of a product = product of limits" and "limit of a sum = sum of limits" hold. The one place it is not automatically legal is a sum whose number of terms grows with $n$ — and Step 6 stops to justify exactly that case rather than assuming it.

Step 1 — One pound, one lump of growth

WHAT. Start with £1. A bank promises 100% growth in one year. If it pays the whole lump once, at year's end you have $start 1 + the 100% growth 1 = 2.$ Here the two $1$ s mean: your original pound, and the extra pound the 100% adds. Nothing subtle yet — this is just "double it."

WHY. We need a baseline. Everything clever later is measured against this £2. If splitting the growth into pieces beats £2, we've found something.

PICTURE (figure s01). In s01 the violet bar on the left is your starting £1; the magenta bar on the right is the £2 you end with. The single orange arrow between them is the one, undivided lump of 100% growth — one jump, no self-feeding yet.

Figure — The number e — definition as limit of (1+1 - n)ⁿ, natural growth context

Step 2 — Split the year into $n$ equal chunks

WHAT. Instead of one lump, chop the year into $n$ equal time-slices. In each slice the bank pays only its share of the 100%, which is $\frac{1}{n}$ of it. So one slice multiplies your money by $1 + \frac{1}{n} .$

The $1$ = "keep what you have."
The $\frac{1}{n}$ = "add this slice's fraction of the 100% growth."

Here $n$ is just a whole number: $n = 2$ means two half-year slices, $n = 12$ means monthly, $n = 365$ means daily.

WHY. Because growth feeds on itself. If you get some growth early, that growth also grows in the later slices. More slices = earlier growth = more self-feeding. This is the entire engine of $e$ .

PICTURE (figure s02). In s02 the single jump of s01 has become a magenta staircase of six treads ( $n = 6$ ). Each little orange arrow lifts you by the same factor $1 + \frac{1}{n}$ ; the violet dots mark your balance after each slice. Notice the treads get taller as you go right — that rising tread height is self-feeding growth made visible.

See Compound interest for the financial version of this same staircase.

Step 3 — Multiply the chunks, don't add them

WHAT. After slice 1 you have $(1 + \frac{1}{n})$ . Slice 2 multiplies that by $(1 + \frac{1}{n})$ again: $after 2 slices = (1 + \frac{1}{n}) \times (1 + \frac{1}{n}) = (1 + \frac{1}{n})^{2} .$ Do this for all $n$ slices and you multiply $n$ times: $year total = (1 + \frac{1}{n})^{n}$

The base $(1 + \frac{1}{n})$ = one slice's multiplier.
The exponent $n$ = how many slices we chain together.

WHY. This is the crux. Compounding is repeated multiplication, never addition. If it were addition you'd get $1 + n \cdot \frac{1}{n} = 2$ every time — no gain from splitting. Multiplication is what lets interest earn interest.

PICTURE (figure s03). In s03 the violet "add" path is a straight line of equal steps that lands flat on the dashed navy line at 2. The magenta "multiply" path curves upward and finishes clearly above that line — the orange arrow points to exactly the extra height that "interest on interest" buys you.

Step 4 — Compute a few values: it climbs but slows

WHAT. Plug in real numbers for $n$ :

Each row uses more slices and gives more money — but the gap between rows shrinks. From $n = 1$ to $n = 2$ we gained $0.25$ ; from $100$ to $1000$ only $0.012$ .

WHY. To feel the two forces fighting: more slices push the value up, but each new slice adds less than the last. We need to know who wins — do we shoot to infinity, or settle down? (This is the first place the limit idea from the opening box bites: we are asking what these values approach.)

PICTURE (figure s04). In s04 the magenta dots are the table's values plotted against $n$ on a log axis. They rise steeply at first, then flatten and press against the dashed violet ceiling line labelled $e \approx 2.71828$ — the orange arrow marks how the gaps keep shrinking toward that line.

Step 5 — Why the sequence keeps rising (never slips back)

WHAT. The table in Step 4 always went up. But "it looked like it went up" is not a proof — we must show that more slices can never give less money. Compare slice-count $n$ against $n + 1$ .

The tool we need is a classic fact about averages:

Apply it with $m = n + 1$ to these $n + 1$ positive numbers (each is $> 0$ since $1 + \frac{1}{n} > 1$ and the spare factor is $1$ ): $n copies (1 + \frac{1}{n}), (1 + \frac{1}{n}), \dots, (1 + \frac{1}{n}), one extra 1 .$

The first $n$ numbers are the slice-multipliers of the $n$ -case.
The last number $1$ is a spare "do nothing" factor we add to make $n + 1$ numbers.

Their arithmetic mean is $\frac{n ( 1 + \frac{1}{n} ) + 1}{n + 1} = \frac{( n + 1 ) + 1}{n + 1} = 1 + \frac{1}{n + 1},$ which is exactly the slice-multiplier of the $n + 1$ -case. AM–GM (all inputs positive, so it applies) gives $1 + \frac{1}{n + 1} \geq n + 1 (1 + \frac{1}{n})^{n} \cdot 1 .$ Raising both sides to the power $n + 1$ (allowed because both sides are positive) gives $(1 + \frac{1}{n + 1})^{n + 1} \geq (1 + \frac{1}{n})^{n} .$

WHY. The left side is the $n + 1$ -case, the right side is the $n$ -case — so each term is at least the one before it. The sequence is monotonically increasing, exactly as the table hinted. This is what turns "it looked like it rose" into a fact we can rely on.

PICTURE (figure s05). In s05 the magenta dots are the sequence values; every orange arrow between consecutive dots points upward and never down, the visual meaning of "monotonically increasing." The navy text callout marks where AM–GM guarantees each dot sits at or above the previous one, all under the dashed violet ceiling $e$ .

Step 6 — Why it doesn't blow up (the factorial brake)

WHAT. Expand $(1 + \frac{1}{n})^{n}$ using the Binomial theorem. The theorem says $(1 + \frac{1}{n})^{n} = \sum_{k = 0}^{n} (k n) \frac{1}{n ^{k}}, (k n) = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - k + 1 )}{k !} .$

$(k n)$ = "how many ways to choose $k$ slices out of $n$ " — the binomial coefficient.
$\frac{1}{n ^{k}}$ = the $k$ factors of $\frac{1}{n}$ that one chosen group contributes.

Now write out term $k$ and split the $n^{k}$ underneath into $k$ separate $n$ 's, pairing each with one factor on top: $(k n) \frac{1}{n ^{k}} = \frac{1}{k !} \cdot k fractions \frac{n}{n} \cdot \frac{n - 1}{n} \cdot \frac{n - 2}{n} \dots \frac{n - k + 1}{n} = \frac{1}{k !} \cdot these are the "messy factors" 1 \cdot (1 - \frac{1}{n}) (1 - \frac{2}{n}) \dots (1 - \frac{k - 1}{n}) .$

Call this term $T_{k} (n) = \frac{1}{k !} \prod_{j = 1}^{k - 1} (1 - \frac{j}{n})$ . There are two separate things to check, because the sum $\sum_{k = 0}^{n} T_{k} (n)$ has a number of terms that grows with $n$ — exactly the case the opening limit box warned us not to swap blindly.

(a) Each fixed- $k$ term tends to $\frac{1}{k}!$ . Hold $k$ still (say $k = 3$ ). Its messy factors are a fixed, finite product $(1 - \frac{1}{n}) (1 - \frac{2}{n})$ ; each bracket $(1 - \frac{j}{n}) \to 1$ as $n \to \infty$ , and a fixed finite product of limits is the product of the limits, so $T_{k} (n) \to \frac{1}{k !}$ . That much is the "safe" term-by-term rule.

(b) The far tail can't sneak in extra mass. We must be sure the growing number of small terms doesn't add up to something. Because every messy factor lies in $(0, 1)$ , we have the clean bound $0 \leq T_{k} (n) \leq \frac{1}{k !} for all n \geq k .$ So the whole sum is squeezed between two limits we can control: $\sum_{k = 0}^{n} T_{k} (n) \leq \sum_{k = 0}^{n} \frac{1}{k !} \leq \sum_{k = 0}^{\infty} \frac{1}{k !} =: S .$ That gives an upper guard. For the lower guard, fix any cutoff $M$ . For $n > M$ , keeping only the first $M + 1$ terms, $\sum_{k = 0}^{n} T_{k} (n) \geq \sum_{k = 0}^{M} T_{k} (n) n \to \infty \sum_{k = 0}^{M} \frac{1}{k !} (a fixed finite sum, part (a)).$ So the limit of the left side is $\geq \sum_{k = 0}^{M} \frac{1}{k !}$ for every $M$ , hence $\geq S$ . Upper guard says $\leq S$ , lower guard says $\geq S$ — squeezed from both sides, the limit is exactly $e = \sum_{k = 0}^{\infty} \frac{1}{k !} = 1 + 1 + \frac{1}{2 !} + \frac{1}{3 !} + \frac{1}{4 !} + \dots$ This is the honest reason the finite $n$ -term sum really converges to the full infinite series, not just term-by-term hand-waving.

Watch the corrections collapse: $\frac{1}{2 !} = 0.5, \frac{1}{3 !} = 0.1 \overline{6}, \frac{1}{4 !} = 0.041 \overline{6}, \frac{1}{5 !} = 0.0083 \dots$

WHY. This is the brake. Factorials grow faster than anything, so each correction is a fraction of the last. For $k \geq 2$ we have $k! \geq 2^{k - 1}$ , so the tail is dwarfed by a halving geometric series: $e = 1 + 1 + \sum_{k \geq 2} \frac{1}{k !} \leq 1 + 1 + \sum_{k \geq 2} \frac{1}{2 ^{k - 1}} = 1 + (1 + \frac{1}{2} + \frac{1}{4} + \dots) = 1 + 2 = 3.$ So the total can never pass 3. Combined with Step 5 (it always rises), a rising sequence with a ceiling must converge — to the number we named $e \approx 2.71828$ .

PICTURE (figure s06). In s06 the coloured left column stacks the terms $\frac{1}{0 !}, \frac{1}{1 !}, \frac{1}{2 !}, \dots$ ; watch how each block is dramatically thinner than the one below — that thinning is the factorial brake. The pale violet right column is the halving comparison $\frac{1}{2 ^{k - 1}}$ , and its total tops out at the dotted magenta line at 3, while the dashed navy line marks where the real stack settles, $e \approx 2.718$ .

Step 7 — The indeterminate tug-of-war (the edge case)

WHAT. Look at the shape of the limit. As $n \to \infty$ :

the base $(1 + \frac{1}{n}) \to 1$ ,
the exponent $n \to \infty$ .

So it's " $1$ raised to $\infty$ ", written $1^{\infty}$ . That is not automatically $1$ .

WHY. Two rival intuitions collide:

"Base is basically $1$ , and $1$ to any power is $1$ " → answer $1$ .
"The power is exploding" → answer $\infty$ .

Both are wrong because they freeze one factor while the other is still moving — precisely the forbidden swap flagged in the opening limit box (here the "number of factors" is the exponent, and it changes with $n$ ). The base is just above 1 (by exactly the amount the exponent needs), and the two effects balance at $e$ , landing strictly between 1 and 3. See Limits and indeterminate forms.

PICTURE (figure s07). In s07 the violet arrow on the left pulls toward the "would-be 1" line; the magenta arrow on the right stretches toward the "would-be ∞" line. The orange circled $e$ sits between them — the outcome of the tug-of-war, pinned at $\approx 2.718$ .

Step 8 — Degenerate & general cases (so nothing surprises you)

WHAT & WHY, case by case.

$n = 1$ (no splitting): $(1 + 1)^{1} = 2$ . The formula still works — it just reproduces Step 1's plain doubling. Good sanity check.
General growth rate $x$ : if the year's total growth is $x$ (not necessarily $100%$ ), each slice pays $\frac{x}{n}$ , so $e^{x} = lim_{n \to \infty} (1 + \frac{x}{n})^{n} .$ Setting $x = 1$ recovers plain $e$ . Here $x$ can be any real number.
$x = 0$ (no growth): $(1 + 0)^{n} = 1$ for all $n$ , and indeed $e^{0} = 1$ . Nothing grows — correct.
$x < 0$ (decay): the base $1 + \frac{x}{n}$ dips below 1, so the product shrinks toward $e^{x} < 1$ . This is exactly natural decay $\frac{dy}{dt}=ky$ with $k<0$. A subtlety worth stating. For small $n$ the base $1 + \frac{x}{n}$ can be zero or negative — e.g. $x = - 2, n = 1$ gives base $- 1$ ; $x = - 2, n = 2$ gives base $0$ . A negative base raised to a whole power still computes, but it can jump sign and is not the growth picture we want. This does not harm the limit: once $n > ∣ x ∣$ we have $\frac{∣ x ∣}{n} < 1$ , so the base $1 + \frac{x}{n} > 0$ and stays positive for all larger $n$ . Since a limit only cares about the eventual tail (all sufficiently large $n$ ), those few early misbehaving terms are irrelevant, and $lim_{n \to \infty} (1 + \frac{x}{n})^{n} = e^{x}$ is well-defined for every real $x$ .

PICTURE (figure s08). In s08 three curves rise or fall against $n$ : the magenta one ( $x = 1$ ) climbs to its dashed line at $e$ ; the violet one ( $x = 0$ ) runs flat along $1$ ; the orange one ( $x = - 1$ ) sinks toward its dashed line at $e^{- 1} \approx 0.368$ . Each curve is only plotted where the base is positive, matching the subtlety above.

The one-picture summary

WHAT. Everything at once: the staircase always rising (Step 5), the shrinking factorial blocks capping it below 3 (Step 6), the tug-of-war balancing base against exponent (Step 7) — all funnelling to the single dashed line at $e \approx 2.71828$ .

PICTURE (figure s09). In s09 the magenta rising dots are the sequence; the shaded band under them shows how far it has climbed. The dotted orange line at 3 is the factorial-brake ceiling from Step 6, and the dashed violet line is the limit $e$ the dots press against. The two annotations tie the two guarantees together: "always rises" (Step 5) and "capped below 3" (Step 6) — a rising thing under a ceiling must land, and it lands on $e$ .

Recall Feynman retelling (open after attempting)

You have £1 in a magic bank that doubles money in a year. Greedy, you ask for the growth in pieces so early growth can itself grow. Two half-year pieces → £2.25. Monthly → £2.61. Every instant → you'd expect a fortune. Two facts pin it down: (1) more pieces always gives more, never less — the sequence only ever rises (that's the average-beats-the-root fact, AM–GM); (2) each new piece adds a tinier bit than the last, because the corrections are $\frac{1}{k !}$ and factorials explode, so the total can never climb past 3. A thing that always rises but is trapped under a ceiling must settle somewhere — and that somewhere is about £2.72, the number we call $e$ . The "base almost 1 fighting an exploding power" is why the answer isn't 1 and isn't infinity — it's this special in-between number. A limit, remember, just means the value these get and stay close to.

Recall Quick checks

Value at $n = 1$ ? ::: $2$ Why is the sequence guaranteed to rise (not just "looks like it")? ::: AM–GM (average $\geq$ geometric mean of positive numbers) shows the $n + 1$ term $\geq$ the $n$ term What is $0!$ and why? ::: $0! = 1$ by the empty-product convention (nothing to multiply = 1) Where do the $\frac{1}{k !}$ terms come from? ::: binomial expansion, then the messy factors $(1 - \frac{j}{n}) \to 1$ Why does the finite $n$ -term sum equal the full infinite series? ::: squeeze: each term $\leq \frac{1}{k !}$ (upper guard) and any first- $M$ block $\to \sum_{0}^{M} \frac{1}{k !}$ (lower guard), forcing the limit to $S$ Why bounded above by 3? ::: for $k \geq 2$ , $k! \geq 2^{k - 1}$ , so the tail is under a halving series summing to 1 What kind of indeterminate form is the limit? ::: $1^{\infty}$ Formula with general rate $x$ ? ::: $e^{x} = lim_{n \to \infty} (1 + x / n)^{n}$ For $x < 0$ , why is the limit still fine despite bad early terms? ::: once $n > ∣ x ∣$ the base $1 + x / n > 0$ and stays positive; limits only depend on the tail What is $e^{0}$ from this? ::: $1$ (base is exactly 1 every step)

Connections

Parent: the number $e$
Compound interest — the staircase of Steps 2–3
Binomial theorem — expands the power into the $\sum \frac{1}{k !}$ series
Limits and indeterminate forms — the $1^{\infty}$ tug-of-war and the swap rules
Exponential function e^x and its derivative — where $e^{x}$ goes next
Differential equations dy/dx = ky — the $x < 0$ decay case
Natural logarithm ln x — the inverse that undoes $e^{x}$

$n$ (slices)	$(1 + \frac{1}{n})^{n}$
1	$2.000$
2	$2.250$
12	$2.613$
100	$2.705$
1000	$2.717$
10000	$2.7181$