 Compounding limit for e

Start at one and apply $n$ updates, each multiplying the current value by $1+1/n$. The final value is $(1+1/n{)}^n$: one update gives $2$, two give $2.25$, ten give approximately $2.59374$, and one hundred give approximately $2.70481$. Increasing the number of steps, while reducing each increment proportionally, approaches Euler's number rather than producing unbounded accumulation.

The natural logarithm makes the limit transparent. For $u>0$, its integral definition gives $u/(1+u)\le \ln (1+u)\le u$, by bounding $1/t$ on $[1,1+u]$. With $u=1/n$,The squeeze theorem gives a logarithmic limit of one; continuity of the exponential function gives $(1+1/n{)}^n\to e$. This links a discrete multiplicative rule with continuous exponential growth, without assuming that a finite number of steps equals the limiting value.