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My expanded question:

Is showing $\lim_{z \to \infty} (1+\frac{1}{z})^z$ exists as $z$ goes through real values the same as $\lim_{n \to \infty} (1+\frac{1}{n})^n$ exists as $n$ goes through integer values? If not, how much additional work is needed to make the two equivalent?

I am asking this because I had posted a question which stated a proof that the limit exists for integer values. I just a few minutes ago answered a question which involved the limit as the argument took on real values by linking to my earlier answer. The answer is here: What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?

It then occurred to me that this is not true until it is proved that the limit through integer values is the same as the limit through real values.

The proof through integer values showed that $(1+1/n)^n$ is an increasing sequence and that $(1+1/n)^{n+1}$ is an decreasing sequence, but says nothing about what happens between the integer values.

If it can be shown that $(1+1/z)^z$ is an increasing function of $z$, that would be enough, but I do not know of a proof that is as elementary as the proof that $\lim_{n \to \infty}(1+1/n)^n$ exists. The usual proofs I have seen involve the power series for $\ln(1-z)$. This can be proved in a number of ways, including starting with $\ln'(z) = 1/z$.

I realize that I am meandering, so I'll leave this at this point.

marty cohen
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4 Answers4

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Let's denote $a_n=(1+\dfrac1n)^n$ for the members of the sequence.

Because $0<n\le x<n+1$, then $$ 1<(1+\frac1{n+1})<(1+\frac1x)\le(1+\frac1n). $$ Consequently we get the upper bound $$ (1+\frac1x)^x\le(1+\frac1x)^{n+1}<(1+\frac1n)^{n+1}=a_n\cdot(1+\frac1n), $$ and similarly the lower bound $$ (1+\frac1x)^x\ge(1+\frac1{n+1})^n=a_{n+1}\cdot(1+\frac1{n+1})^{-1}. $$ So $(1+\dfrac1x)^x$ is sandwiched between two consecutive entries of the sequence $(a_n)$ multiplied by terms $\to 1$.

To answer: Yes, the two limit problems are equivalent.

Jyrki Lahtonen
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By the definition of limits.

$c = \lim_{z \to \infty} (1+\frac{1}{z})^z$ exists if and only if $\forall (s_n)_{n\in\mathbb{N}}, \, \, \left(\lim_{n \to \infty}(s_n) = +\infty \right)\implies \left(s = \lim_{n \to \infty} (1+\frac{1}{s_n})^{s_n} \,\mbox{ exists and }s = c\right)$.

corindo
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Let $0 < x_1 < x_2$. Then $0 < \frac{x_1}{x_2} < 1$, so by the Bernoulli's inequality

$$\left( 1 + \frac{1}{x_1} \right)^{x_1} = \left( 1 + \frac{1}{x_1} \right)^{\tfrac{x_1}{x_2} \cdot \ x_2} < \left( 1 + \frac{x_1}{x_2} \cdot \frac{1}{x_1} \right)^{x_2} = \left( 1 + \frac{1}{x_2} \right)^{x_2}.$$

Hence the function $f(x) = \left(1+\frac{1}{x}\right)^x$ is increasing on $(0, \infty)$.

Adayah
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The fundamental difference in dealing with limits $\lim_{n \to \infty}(1 + (1/n))^{n}$ and $\lim_{z \to \infty}(1 + (1/z))^{z}$ is that the conception of the first limit is simpler. The function $f(n) = (1 + (1/n))^{n}$ gives rational values for positive integers $n$ and can be calculated via simple arithmetic.

When you deal with $g(z) = (1 + (1/z))^{z}$ for positive real number $z$, then things are bit complex. The concept of an irrational exponent needs some development. If one has a sound theory of irrational exponents (some approaches are in my blog posts) then the proof that this limit exists is simple (based on taking logs).

In short the limit dealing with $n$ requires just theorems about monotone bounded sequences whereas limit concerning $z$ needs in addition the theory of arbitrary real exponents.

Paramanand Singh
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