Show $\lim\left ( 1+ \frac{1}{n} \right )^n = e$ if $e$ is defined by $\int_1^e \frac{1}{x} dx = 1$

Solution 1:

The proof as stated is circular, but it's easily fixed, more or less by just rephrasing things.

Problem: When you say "Since ... $\lim\left(1+\frac1m\right)^m=e$... we have the required result" you certainly appear to be assuming what you're trying to prove.

Fix: The original inequality shows that $$e\ge\limsup\left(1+\frac1n\right)^n.$$Your manipulations with the original upper bound show that $$e\le\liminf\left(1+\frac1n\right)^n,$$and these two inequalities show that the limit in question exists and equals $e$.

Solution 2:

David C. Ullrich has already given a perfect answer using the tactical device of $\limsup, \liminf$ which helps a lot when the existence of limit is not known in advance.

However since you have used the integral definition for $e$ it is better to use the properties of integrals to prove your claim. So let $$f(x) = \int_{1}^{x}\frac{dt}{t}\tag{1}$$ for $x > 0$. The integral above does not exist if $x \leq 0$. And the definition of $e$ is given by $f(e) = 1$. We have \begin{align} f(xy) &= \int_{1}^{xy}\frac{dt}{t}\notag\\ &= \int_{1}^{x}\frac{dt}{t} + \int_{x}^{xy}\frac{dt}{t}\notag\\ &= f(x) + \int_{1}^{y}\frac{d(vx)}{vx}\text{ (putting }t = vx)\notag\\ &= f(x) + \int_{1}^{y}\frac{dv}{v}\notag\\ &= f(x) + f(y)\notag \end{align} Using the above property repeatedly we see that $f(x^{n}) = nf(x)$ for positive integer $n$ and $x > 0$. Therefore if $a_{n} = \left(1 + \dfrac{1}{n}\right)^{n}$ then $$f(a_{n}) = nf\left(1 + \frac{1}{n}\right)\tag{2}$$ We will show that that the RHS of the above equation tends to $1$ and therefore LHS also does the same.

If $1 < t < 1 + 1/n$ then $$\frac{n}{n + 1} < \frac{1}{t} < 1$$ and upon integrating this inequality on interval $[1, 1 + 1/n]$ we get $$\frac{1}{n + 1} < f\left(1 + \frac{1}{n}\right) < \frac{1}{n}$$ or $$\frac{n}{n + 1} < nf\left(1 + \frac{1}{n}\right) < 1$$ and by Squeeze theorem we now see that $$\lim_{n \to \infty}nf\left(1 + \frac{1}{n}\right) = 1$$ and therefore from $(2)$ we get $$\lim_{n \to \infty}f(a_{n}) = 1 = f(e)\tag{3}$$ Now a quick but incorrect route to the answer is to use the continuity of $f$ to replace $\lim f(a_{n})$ by $f(\lim a_{n})$ to get $f(\lim a_{n}) = f(e)$ and then use one-one property of $f$ to get $\lim a_{n} = e$. But the problem with this approach is that we don't know whether limit of $a_{n}$ exists or not (BTW the existence of this limit is routinely proved in many real-analysis/calculus textbooks but we don't use that approach here).

Instead we note that $f(x)$ is differentiable for $x > 0$ with a positive derivative and therefore there exists a unique inverse $g$ of $f$ which is also differentiable with positive derivative. The domain/range of $g$ match the range/domain of $f$ and it can be proved that $f$ maps $\mathbb{R}^{+}$ to $\mathbb{R}$ so that the inverse $g$ maps $\mathbb{R}$ to $\mathbb{R}^{+}$. Applying $g$ on equation $(3)$ we get $$g(\lim_{n \to \infty}f(a_{n})) = g(f(e)) = e$$ and by continuity of $g$ we get $$\lim_{n \to \infty}g(f(a_{n})) = e$$ or $$\lim_{n \to \infty}a_{n} = e$$ which we set out to prove.

It should be obvious to many readers that the functions $f, g$ are $\log$ and $\exp$ respectively, but it is better to use the symbols $f, g$ in order to avoid using any well known properties of $\log, \exp$. This is a great help in avoiding circular proofs.