What happens to (1 + 1/n)ⁿ as n grows?

A bank pays interest at a nominal annual rate of 100%. If it compounds more and more often, monthly, then daily, then every second, how much does $1 grow to in one year?

Compounding n times a year multiplies the balance by 1 + 1/n, n times over. More frequent compounding pays interest on interest sooner, but each step is smaller. Stage 1 asks you to predict where the year-end balance goes as n grows without bound.

Before you start · 3 quick checks on compound interest, exponent rules and limits of sequences, about 3 min

Definitions

Compounding. A nominal annual rate $r$ compounded $n$ times a year means that the year is cut into $n$ equal periods, and at the end of each period the bank adds $\tfrac{r}{n}$ times the current balance. So each period multiplies the balance by $1 + \tfrac{r}{n}$, and after one year a deposit $P$ has become

\[ P\Big(1 + \frac{r}{n}\Big)^{n}. \]

The rate is written as a decimal: 6% a year is $r = 0.06$, and 100% a year is $r = 1$.

Limit of a sequence. A sequence $s_1, s_2, s_3, \ldots$ tends to the number $L$, written $s_n \to L$ or $\lim_{n \to \infty} s_n = L$, when for every $\varepsilon \gt 0$ there is an $N$ with $|s_n - L| \lt \varepsilon$ for every $n \ge N$. It grows without bound, written $s_n \to \infty$, when for every number $B$ there is an $N$ with $s_n \gt B$ for every $n \ge N$.

Fine print: nominal rate and growth in one year

Banks quote the nominal rate $r$. The actual growth over one year, $(1 + \tfrac{r}{n})^n - 1$ as a fraction of the deposit, depends on how often the interest is compounded. On this page there are no fees, taxes, deposits or withdrawals beyond those stated, and the rate stays fixed.

1 See

Predict the year-end balance, then watch the staircase

One click: where do you think the year-end balance goes? After you answer, a figure shows $1 growing during the year, one step up at each compounding, and you can set the number of steps $n$.

The six compounding frequencies shown above. A value marked = is exact; a value marked ≈ is rounded to the digits shown.
n	Each step: 1 + 1/n	Year end: (1 + 1/n)ⁿ	Below e by

How the numbers are computed: for large $n$ the page computes $(1 + \tfrac{1}{n})^n$ as $e^{n\ln(1 + 1/n)}$, with $\ln(1 + u)$ evaluated directly for small $u$, instead of multiplying $n$ rounded factors. This matters: in the double-precision arithmetic of a computer, for $n = 10^{16}$ the number $1 + \tfrac{1}{n}$ rounds to exactly 1, and multiplying it by itself $n$ times would return 1.

2 Try

Other rates: (1 + x/n)^n

Write the annual rate as a decimal $x$, so $x = 1$ is 100% a year. Compounded $n$ times a year, $1 becomes $(1 + \tfrac{x}{n})^n$ dollars after one year. A negative $x$, with $n \gt |x|$ so that the base $1 + \tfrac{x}{n}$ is positive, describes a quantity that shrinks.

The explorer and its goals open after you answer the prediction.

3 Prove

Why the limit is e^x

What the proof uses

The number $e$. The sequence $(1 + \tfrac{1}{n})^n$ increases with $n$, and every term is below 3 (optional proof below). An increasing sequence that stays below a fixed number tends to a limit, so the limit exists; it is called $e$: \[ e = \lim_{n \to \infty}\Big(1 + \frac{1}{n}\Big)^{n} \approx 2.718281828. \] Since the terms increase, every term is below $e$.
Exponential and logarithm. For real $t$, the number $e^t$ is positive, and $t \mapsto e^t$ is continuous and increasing, with $e^{s + t} = e^s e^t$ and $e^0 = 1$. Its inverse is the natural logarithm: for $y \gt 0$, $\ln y$ is the number with $e^{\ln y} = y$, and $\ln(e^t) = t$ for every real $t$. The logarithm is continuous on $(0, \infty)$, and $\ln(u^n) = n\ln u$ for $u \gt 0$ and every integer $n$.
The slope of $\ln$ at 1. $\displaystyle\lim_{h \to 0} \frac{\ln(1 + h)}{h} = 1$, where $h$ tends to 0 through nonzero values with $h \gt -1$. Since $\ln 1 = 0$, the quotient is $\tfrac{\ln(1 + h) - \ln 1}{h}$, so this says $\ln'(1) = 1$. The box below derives it from the definition of $e$.

Optional: why (1 + 1/n)^n increases and stays below 3

It increases. The arithmetic and geometric mean inequality says that for positive numbers $c_1, \ldots, c_m$, not all equal, $(c_1 c_2 \cdots c_m)^{1/m} \lt \tfrac{c_1 + \cdots + c_m}{m}$. Take $m = n + 1$ numbers: $n$ copies of $1 + \tfrac{1}{n}$ and one copy of 1. Their sum is $n + 1 + 1 = n + 2$, so

\[ \Big(1 + \frac{1}{n}\Big)^{\frac{n}{n+1}} \lt \frac{n + 2}{n + 1} = 1 + \frac{1}{n + 1}. \]

Raising both positive sides to the power $n + 1$ gives $(1 + \tfrac{1}{n})^n \lt (1 + \tfrac{1}{n+1})^{n+1}$ for every $n \ge 1$.

It stays below 3. By the binomial theorem, $(1 + \tfrac{1}{n})^n = \sum_{k=0}^{n} \binom{n}{k}\tfrac{1}{n^k}$, and for $k \ge 1$

\[\begin{aligned} &\binom{n}{k}\frac{1}{n^k} \\ &= \frac{1}{k!}\cdot\frac{n(n-1)\cdots(n-k+1)}{n^k} \\ &\le \frac{1}{k!}, \end{aligned}\]

because each of the $k$ factors $n, n - 1, \ldots, n - k + 1$ is at most $n$.

Since $2^{k-1} \le k!$ for $k \ge 1$, the sum is at most $1 + \big(1 + \tfrac{1}{2} + \tfrac{1}{4} + \cdots + \tfrac{1}{2^{n-1}}\big) \lt 1 + 2 = 3$.

Why ln′(1) = 1 follows from the definition of e

From the right. Let $0 \lt h \le 1$ and let $m$ be the whole number with $m \le \tfrac{1}{h} \lt m + 1$, so $\tfrac{1}{m+1} \lt h \le \tfrac{1}{m}$ and $m \ge 1$. A larger base and a larger exponent (with base above 1) give a larger power, so

\[\begin{aligned} \Big(1 + \frac{1}{m+1}\Big)^{m} &\le (1 + h)^{1/h} \\ &\le \Big(1 + \frac{1}{m}\Big)^{m+1}. \end{aligned}\]

As $h \to 0^+$, $m$ grows without bound. The right side is $(1 + \tfrac{1}{m})^m$ times $1 + \tfrac{1}{m}$, and the left side is $(1 + \tfrac{1}{m+1})^{m+1}$ divided by $1 + \tfrac{1}{m+1}$; both tend to $e$. So $(1 + h)^{1/h} \to e$ as $h \to 0^+$.

From the left. For $-1 \lt h \lt 0$, write $h = -k$ with $0 \lt k \lt 1$ and put $u = \tfrac{k}{1 - k} \gt 0$. Then $\tfrac{1}{1 - k} = 1 + u$ and $\tfrac{1}{k} = \tfrac{1}{u} + 1$, so $(1 - k)^{-1/k} = (1 + u)^{1/u}\,(1 + u)$. As $k \to 0^+$, $u \to 0^+$, and this tends to $e \cdot 1 = e$.

Taking logarithms. So $(1 + h)^{1/h} \to e$ as $h \to 0$ from either side. Since $\ln$ is continuous at $e$, $\tfrac{\ln(1 + h)}{h} = \ln\big((1 + h)^{1/h}\big) \to \ln e = 1$.

Theorem

For every real number $x$, with $a_n = (1 + \tfrac{x}{n})^n$ for integers $n \ge 1$,

\[ \lim_{n \to \infty}\Big(1 + \frac{x}{n}\Big)^{n} = e^{x}. \]

Consequences

Continuous compounding. At a nominal rate $r$ compounded $n$ times a year, after $t$ whole years a deposit $P$ has become $P(1 + \tfrac{r}{n})^{nt} = P\big((1 + \tfrac{r}{n})^n\big)^t$. As $n \to \infty$ this tends to $P(e^r)^t = Pe^{rt}$, the balance under continuous compounding.

The derivative of $e^x$ is $e^x$. For $h \ne 0$, $\dfrac{e^{x+h} - e^x}{h} = e^x\,\dfrac{e^h - 1}{h}$. Put $u = e^h - 1$. Then $u \ne 0$, $u \to 0$ as $h \to 0$, and $h = \ln(1 + u)$, so $\dfrac{e^h - 1}{h} = \dfrac{u}{\ln(1 + u)} \to 1$ by the slope of $\ln$ at 1. So $(e^x)' = e^x$, and by the chain rule $(e^{kx})' = k\,e^{kx}$: a continuously compounded balance $Pe^{rt}$ changes at the rate $r$ times itself: it grows when $r \gt 0$ and shrinks when $r \lt 0$.

The derivative of $\ln x$ is $\tfrac{1}{x}$. For $x \gt 0$ and $h \ne 0$ with $x + h \gt 0$, $\ln(x + h) - \ln x = \ln(1 + \tfrac{h}{x})$. With $k = \tfrac{h}{x}$, which is nonzero and tends to 0 as $h \to 0$,

\[\begin{aligned} \frac{\ln(x + h) - \ln x}{h} &= \frac{1}{x}\cdot\frac{\ln(1 + k)}{k}, \end{aligned}\]

which tends to $\tfrac{1}{x}$ as $h \to 0$. Once you know that $\ln$ is differentiable, the chain rule gives the same answer from $e^{\ln x} = x$: $e^{\ln x}\,(\ln x)' = 1$, so $(\ln x)' = \tfrac{1}{e^{\ln x}} = \tfrac{1}{x}$. Another common route defines $\ln x = \int_1^x \tfrac{1}{t}\,dt$; then the fundamental theorem of calculus gives $(\ln x)' = \tfrac{1}{x}$ at once.

4 Apply

Problems

Each problem has hints and a full solution. Money answers are in dollars, with no dollar sign needed.

What happens to (1 + 1/n)ⁿ as n grows?

Predict the year-end balance, then watch the staircase

Other rates: (1 + x/n)^n

Goals: how fast, and where it breaks

Why the limit is e^x

Problems

Further reading

What happens to (1 + 1/n)n as n grows?

Predict the year-end balance, then watch the staircase

Other rates: (1 + x/n)^n

Goals: how fast, and where it breaks

Why the limit is e^x

Problems

Further reading

What happens to (1 + 1/n)ⁿ as n grows?