Is showing $\lim_{z \to \infty} (1+\frac{1}{z})^z$ exists the same as $\lim_{n \to \infty} (1+1/n)^n$ exists

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.

Solution 1:

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.