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The log-free product limit $(1-2/n)^n\to\exp(-2)$

Example

Define a sequence $(a_n)_{n\in\mathbb N}$ by $a_0:=0,\qquad a_n:=\left(1-\frac2{\iota(n)}\right)^n\quad(n\ge1).$ Then $a_n\longrightarrow e^{-2}$ . The separate value at $n=0$ avoids division by $\iota(0)=0$ ; the finitely many remaining initial indices with nonpositive base do not affect the limit.

Facts & Assumptions

Given: The sequence $(a_n)$ defined above.

[L1]

The product-limit theorem holds for every real input once $n>|x|$ (For every real $x$ , $(1+x/n)^n\to\exp x$ ).

[L2]

Since $e=\exp(1)$ , the addition and reciprocal formulas and the definition of negative integer powers give $\exp(-2)=1/\exp(2)=1/(\exp(1)\exp(1))=1/e^2=e^{-2}>0$ (The real exponential function and the number $e$ by a power series, The exponential addition formula $\exp(x+y)=\exp(x)\exp(y)$ , The exponential is positive and satisfies $\exp(-x)=1/\exp(x)$ , Integer powers $a^m$ ).

[L3]

A sequence and any one of its tails have the same limit (Convergence depends only on the tail).

Verification

technique · direct

1.1

For every $n>2$ , the base is positive and $a_n=(1-2/\iota(n))^n$ , so [L1] applies at $x=-2$ and gives $a_n\to\exp(-2)$ on that tail.

2.1

By [L3], the whole sequence has the same limit as the tail in step 1.1, and [L2] identifies that limit as $e^{-2}>0$ .

step 1.1L2L3∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 93 results over 19 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)

The log-free product limit (1−2/n)n→exp⁡(−2)(1-2/n)^n\to\exp(-2)(1−2/n)n→exp(−2)