Show that $(1+p/n)^n$ is a Cauchy sequence for arbitrary $p$ It is a generalization of this question. I am looking for a similar derivation as in here.
Can we prove that $(1+p/n)^n$ is a Cauchy sequence for any $p \in [a, b]$ by showing that
$$ \Bigg| \left( 1 + \frac{p}{n}\right)^n - \left( 1 + \frac{p}{m}\right)^m \Bigg| \leq f(n)$$
where $f(n)$ is something that tends to zero as $n$ goes to infinity?
Here is an attempt.
Let $m=n+1$. Then,
$$ \Bigg| \left( 1 + \frac{p}{n}\right)^n - \left( 1 + \frac{p}{n+1}\right)^{n+1} \Bigg| = \Bigg| \sum_{k=0}^{n} \binom{n}{k}\left( \frac{p}{n} \right)^{k} - \sum_{k=0}^{n+1} \binom{n+1}{k}\left( \frac{p}{n+1} \right)^{k} \Bigg| = \\
\Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \left[ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) \right] + \frac{p^{n+1}}{(n+1)^{n+1}} \Bigg| \leq
\sum_{k=0}^{n} \frac{|p|^k}{k!} \left[ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) \right] + \Bigg| \frac{p^{n+1}}{(n+1)^{n+1}} \Bigg|
$$
Now intuitively, $\sum_{k=0}^{n} \frac{|p|^k}{k!}$ tends to the constant $e^{|p|}$ while the expression in square brackets tends to zero as $n$ goes to infinity. I was trying to show that this expression is less than a constant divided by $n^2$ (because we will sum up the consecutive terms for $m >n$, and the sum should converge which $\frac{C}{n^2}$ would provide).
Can we also find an estimate of the constant $C$?

Thanks to all for good answers. Meanwhile, I'm still quite interested in working it out algebraically. My next idea was to use the triangle inequality. Let us ignore the last term in the inequality above, it obviously tends to zero as $n$ goes to infinity. We want the sum to go to zero fast enough. So,
$$ \Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) \Bigg| \leq \\
\Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) - \sum_{k=0}^{n} \frac{p^k}{k!} \Bigg| + \\
\Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \sum_{k=0}^{n} \frac{p^k}{k!} \Bigg|
$$
Each product is evidently positive and smaller than one. So we end up having kind of a usual exponential series, but weighted. Otherwise, we could try to plug something instead of the products, what would, when subtracted from the products, give something which decreases fast enough. I feel this should be possible since we have all the freedom to plug in whatever we want.

Sequence is Cauchy
Bernoulli's Inequality shows that the sequence is monotonically increasing.
$$
\begin{align}
\left.\left(1+\frac p{n+1}\right)^{n+1}\middle/\left(1+\frac pn\right)^n\right.
&=\frac{n+p}{n}\left(\frac{n+p+1}{n+1}\frac{n}{n+p}\right)^{n+1}\\
&=\frac{n+p}{n}\left(1-\frac{p}{(n+1)(n+p)}\right)^{n+1}\\
&\ge\frac{n+p}{n}\left(1-\frac{p}{n+p}\right)\\[4pt]
&=1\tag{1}
\end{align}
$$
If $p\le0$, then $\left(1+\frac pn\right)^n\le1$, so $\left(1+\frac pn\right)^n$ is bounded above.
If $p\ge0$, the Binomial Theorem shows that the sequence is bounded above.
$$
\begin{align}
\left(1+\frac pn\right)^n
&=1+\frac n1\frac pn+\frac{n(n-1)}{2!}\left(\frac pn\right)^2+\frac{n(n-1)(n-2)}{3!}\left(\frac pn\right)^3+\cdots\\
&=1+\frac nn\frac p1+\frac{n(n-1)}{n^2}\frac{p^2}{2!}+\frac{n(n-1)(n-2)}{n^3}\frac{p^3}{3!}+\cdots\\
&\le1+p+\frac{p^2}{2!}+\frac{p^3}{3!}+\cdots\\[6pt]
&=e^p\tag{2}
\end{align}
$$

Since $\left(1+\frac pn\right)^n$ is increasing and bounded above, it is convergent, hence Cauchy.

If we are willing to use the inequality $1+x\le e^x$, then we can simplify and extend the proof of $(2)$ to $1+\frac pn\le e^{p/n}$ and raise both sides to the $n^{\text{th}}$ power. This shows that $\left(1+\frac pn\right)^n\le e^p$ for all $p\gt-n$ (and since we are looking at large $n$, this is fine).

Regarding $\ \boldsymbol{f}$
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[\log(1+x)-x+\frac{x^2}2\right]
=\frac{x^2}{1+x}
\ge0\tag{3}
$$
Thus, for $x\ge0$,
$$
\log(1+x)\ge x-\frac{x^2}2\tag{4}
$$
Therefore, for $p\ge0$, we have
$$
n\log\left(1+\frac pn\right)\ge p-\frac{p^2}{2n}\tag{5}
$$
Thus,
$$
\begin{align}
e^p
&\ge\left(1+\frac pn\right)^n\tag{6}\\
&\ge e^pe^{-\frac{p^2}{2n}}\tag{7}\\[3pt]
&\ge e^p\left(1-\frac{p^2}{2n}\right)\tag{8}
\end{align}
$$
Explanation:
$(6)$: apply $(2)$
$(7)$: apply $\exp$ to $(5)$
$(8)$: $e^x\ge1+x$
Thus, for $m\gt n$, we have
$$
\begin{align}
\left(1+\frac pm\right)^m-\left(1+\frac pn\right)^n
&\le e^p-\left(1+\frac pn\right)^n\\
&\le e^p\frac{p^2}{2n}\tag{9}
\end{align}
$$
Therefore,
$$
f(n)=e^p\frac{p^2}{2n}\tag{10}
$$
works.
