Proof, without logs/exp that (1+x/n)^n converges

Question

was wondering if any of you would be able to help with this question:

Prove that $$\left(1 +\frac{x}{n}\right)^{n}$$ converges as n approaches infinity for $x\in\mathbb{R}$ and $n=1,2,3...$

Now, since this question was given before we learnt the proper definition of exponential functions and logarithms, I assume it requires a proof without these functions. I was able to prove the convergence for $x\geq0$ using the binomial expansion and the fact that $$\binom{n+1}{k}\frac{x^{k}}{(n+1)^{k}}>\binom{n}{k}\frac{x^{k}}{n^{k}}$$ for $x\geq0$ (i.e I proved the sequence to be increasing and was bounded above, and hence convergent). However, that inequality is the other way round for $x<0$ and the inequalities I used to prove the sequence to be bounded above also relied on the fact that $x\geq0$.

How would I go about proving that the sequence is increasing and bounded above (without logs/exponentials) for $x<0$? Would I have to come up with some new inequalities?

Thanks

Answer 1

Instead of proving that the sequence is convergent for negative $x$, let us prove instead that $$\left(1 +\frac{x}{n}\right)^{n}\left(1 -\frac{x}{n}\right)^{n}=\left(1 -\frac{x^2}{n^2}\right)^{n}$$ converges to $1$.

Indeed, for all $n > |x|$ we have by Bernoulli $$1\geq \left(1 -\frac{x^2}{n^2}\right)^{n} \geq 1-n\frac{x^2}{n^2}$$

Therefore squeeze shows the claim.

Answer 2

In THIS ANSWER, I showed using Bernoulli's Inequality that the sequence $e_n(x)=\left(1+\frac xn\right)^n$ monotonically increases for $x>-n$.

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CASE $1$: $\displaystyle x\le 0$

Note that if $-n<x\le 0$, then $0< e_n(x)\le 1$. So, $e_n(x)$ monotonically increases (for $x>-n$) and is bounded above by $1$. Hence $e_n(x)$ converges for $x\le 0$.

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CASE $2$: $\displaystyle x> 0$

To show that the sequence is bounded above for $x>0$, we write for $x>-n$

$$\begin{align} 0&\le \left(1+\frac xn\right)^n\\\\ &=\sum_{k=0}^n\binom{n}{k}\left(\frac{x}{n}\right)^k\\\\ &=\sum_{k=0}^n\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)\,\frac{x^k}{k!}\\\\ &\le \sum_{k=0}^\infty \frac{x^k}{k!} \end{align}$$

which converges for all $x$ (use the root test or ratio test to see this).

Therefore, for any $x>0$, $e_n(x)$ is monotonically increasing for all $n$ and bounded above by $\sum_{n=0}^\infty \frac{x^n}{n!}$. Hence, $e_n(x)$ converges.