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For every real $x$ , $(1+x/n)^n\to\exp x$

Statement

For every real $x$ , $\lim_{n\to\infty}\left(1+\frac{x}{\iota(n)}\right)^n=\exp(x),$ with the sequence started after $n>|x|$ , so the base is positive.

Facts & Assumptions

Given: A real $x$ .

[L1]

The binomial theorem expands the product. For fixed $k$ , For fixed $k$ , $\binom{n}{k}/n^k$ tends to $1/k!$ gives both convergence of the scaled coefficient to $1/\iota(k!)$ and, whenever $n\ge\max\{1,k\}$ , the bound $0\le\iota\binom nk/\iota(n)^k\le1/\iota(k!)$ (The binomial theorem in $\mathbb{R}$ : $(x+y)^{n} = \sum_{k<n+1} \iota\!\binom{n}{k}\, x^{k} y^{\,n-k}$ ).

[L2]

The exponential series converges absolutely (The exponential series converges absolutely for every real argument, The real exponential function and the number $e$ by a power series).

Proof

technique · direct

1.1

For $n>|x|$ , the binomial theorem gives $(1+x/\iota(n))^n=\sum_{k=0}^{n}\bigl(\iota\binom nk/\iota(n)^k\bigr)x^k$ .

1.2

Each fixed coefficient tends to $1/\iota(k!)$ , while the uniform bound in [L1] holds for every term present in the sum.

2.1

Given $\varepsilon>0$ , choose $K$ so the absolute exponential tail after $K$ is below $\varepsilon/3$ using [L2]. The same coefficient bound controls the product tail uniformly in $n$ ; for the finite head $k\le K$ , choose $n$ so all coefficient errors sum to below $\varepsilon/3$ .

step 1.2L2choose

3.1

The triangle inequality then makes the product differ from $\sum_{k\ge0}x^k/\iota(k!)=\exp(x)$ by less than $\varepsilon$ .

step 1.1step 2.1given∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 102 results over 26 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)
J. Lebl, Basic Analysis, Analytic Functions (standard reference, not scraped)
J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)

For every real xxx, (1+x/n)n→exp⁡x(1+x/n)^n\to\exp x(1+x/n)n→expx