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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.

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    $\begingroup$ $(1 + a/n)^n$ converges to $e^a$ for any $a$. A convergent sequence is a Cauchy sequence. $\endgroup$ Jul 29 '15 at 18:25
  • $\begingroup$ Ok. I will try to reformulate $\endgroup$
    – Rubi Shnol
    Jul 29 '15 at 18:26
  • $\begingroup$ @RaceBannon: The limit of this sequence is one of the many equivalent definitions of $e^a$. In order for the definition to make sense, it must be established that these sequences converge, without already having a definition for what they converge to, so either Cauchiness or monotonicity and boundedness are proven. I suspect that this is the exercise, and using the fact that it converges to $e^a$ is, for lack of a better word, cheating. :-) $\endgroup$ Jul 29 '15 at 18:31
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    $\begingroup$ @ValerySaharov: When proving the Cauchiness of a sequence $(a_n)$, it is not sufficient to simply prove $|a_n - a_{n+1}| \rightarrow 0$ as $n \rightarrow \infty$. You cannot simply choose $m$ based on $n$! Consider the partial sums of the harmonic series as a counterexample. $\endgroup$ Jul 29 '15 at 18:33
  • $\begingroup$ Yes, that would work. If $|a_n - a_{n+1}| < K/n$ eventually for some $K$, then $$a_n = a_1 + \sum_{k=1}^{n-1} a_{k+1} - a_k$$ is an absolutely convergent series. $\endgroup$ Jul 29 '15 at 18:42
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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.

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  • $\begingroup$ @ValerySaharov: If $p\le0$, then $1+p\le\left(1+\frac pn\right)^n\le1$, so the sequence is bounded and increasing, therefore, Cauchy. We don't need the second part to show boundedness. $\endgroup$
    – robjohn
    Jul 30 '15 at 12:12
  • $\begingroup$ @ValerySaharov: No, my inequality is supposed to be $$\left(1+\frac pn\right)^n \le\sum_{k=0}^n\binom{n}{k}\frac{p^k}{n^k} \le\sum_{k=0}^\infty\frac{p^k}{k!}$$ for $p\ge0$. We could actually take the upper limit in the middle sum to be $\infty$ since $\binom{n}{k}=0$ when $k\gt n$. $\endgroup$
    – robjohn
    Jul 30 '15 at 12:17
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One approach using L'Hospital to show $$\lim\limits_{x\to\infty}\left(1+\frac{a}{x}\right)^x=\exp(a).$$ This will prove that the sequence is a cauchy sequence.

Let $$F(x)=x\ln\left(1+\frac{a}{x}\right)=\frac{\ln\left(1+\frac{a}{x}\right)}{\frac{1}{x}}.$$ We have $\lim\limits_{x\to\infty}\ln\left(1+\frac{a}{x}\right)=0=\lim\limits_{x\to\infty}\frac{1}{x}$, thus by applying L'Hospitals rule (and assuming we checked that the other conditions are given) we get $$\lim\limits_{x\to\infty} F(x)=\lim\limits_{x\to\infty}\frac{\frac{1}{1+a/x}\cdot\frac{-a}{x^2}}{-\frac{1}{x^2}}=\lim\limits_{x\to\infty}\frac{a}{1+\frac{a}{x}}=a.$$

As $\left(1+\frac{a}{x}\right)^x=\exp(F(x))$ and $\exp$ is continuous we get: $$\lim\limits_{x\to\infty}\left(1+\frac{a}{x}\right)^x=\exp\left(\lim\limits_{x\to\infty}F(x)\right)=\exp(a).$$

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  • $\begingroup$ I suspect that the reason the OP wants to show it is a Cauchy sequence is in order to develop the exponential in the first place. $\endgroup$
    – 6005
    Jul 29 '15 at 18:49
  • $\begingroup$ Well, the request for a specific way was edited in later, but I guess if you don't know $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e$ yet, you'd have to prove that first. $\endgroup$
    – Hirshy
    Jul 29 '15 at 18:56
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Here is a sketch of proving that $f(n)$ is Cauchy. Whenever we want to bound $|f(a)-f(b)|$ then the mean value theorem is usually our best friend.

The mean value theorem says that for a differentiable function $f$ we have $$|f(n) - f(m)| = |f'(c)||n-m| \text{ for some } c\in(n,m)$$

For this case $f(x) = \left(1+\frac{a}{x}\right)^x$. We first need to bound the derivative

$$f'(x) = f(x)\left[\log\left(1+\frac{a}{x}\right) - \frac{a}{x+a}\right]$$

For large enough $x$ we find by Taylor expanding that $f'(x) \simeq f(x) \frac{a^2}{2x^2}$. Next we can show that $f(x)$ is bounded on $[0,\infty)$. Putting all the ingredients togeather it follows that for $n<m$ large enough

$$|f(n) - f(m)| \leq \frac{C}{n^2}|n-m|$$

for some constant $C$ independent of $n$. There is a lot of missing details above, but those should be fairly straight forward to fill in.

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  • $\begingroup$ @ValerySaharov Unfortunately. Your approach might work out, but I suspect the algebra (when estimating the different terms) might become very messy. Thats the main reason why I suggested using the MVT instead: you only need some very simple estimates to get to the result you want. $\endgroup$
    – Winther
    Jul 29 '15 at 19:33
  • $\begingroup$ @ValerySaharov You could try to use a Taylor expansion directly instead of working with the exact sum. We have $(1+a/n)^n = e^a\left(1 - \frac{a^2}{2n} + \mathcal{O}(1/n^2)\right)$ so $|f(n) - f(m)| = \frac{e^aa^2}{2}\left|\frac{n-m}{nm}\right| + \mathcal{O}(1/n^2) + \mathcal{O}(1/m^2)$. $\endgroup$
    – Winther
    Jul 29 '15 at 19:52
  • $\begingroup$ I have added an answer that proves the estimate for $\ln(1+z)$. It should be a comment, but that would be a real pain. $\endgroup$ Jul 30 '15 at 5:24
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This shows the estimate needed in Winther's answer. It is too complicated to be a comment, though that is what it really it.

Define $\ln(1+x) =\int_1^{1+x} \frac{dt}{t} $. Since, for $1 \le t \le 1+x$ $\frac{1}{1+x} \le \frac{1}{t} \le 1$, $\frac{x}{1+x} \le \ln(1+x) =\int_1^{1+x} \frac{dt}{t} \le x $, so that $\frac{x}{1+x}-x \le \ln(1+x)-x \le 0 $ or, reversing, $0 \le x-\ln(1+x) \le \frac{x^2}{1+x} $ .

Therefore $0 \le \frac{a}{x}-\ln(1+a/x) \le \frac{(a/x)^2}{1+a/x} =\frac{a^2}{x(x+a)} $.

Since $\frac{a}{x}-\frac{a}{x+a} =\frac{a^2}{x(x+a)} $, $0 \le \frac{a}{x+a}+\frac{a^2}{x(x+a)}-\ln(1+a/x) \le \frac{a^2}{x(x+a)} $ or $0 \le \ln(1+a/x)-\frac{a}{x+a} \le \frac{a^2}{x(x+a)} \le \frac{a^2}{x^2} $.

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  • $\begingroup$ I missed the point a little bit. This is now Winther's $C$? $\endgroup$
    – Rubi Shnol
    Jul 30 '15 at 6:30
  • $\begingroup$ $a^2$ is the $C$, $x^2$ is $n^2$. $\endgroup$ Jul 30 '15 at 21:53
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Proving directly that $a_{n} = (1 + (p/n))^{n}$ is a Cauchy sequence seems difficult especially when one wants to use only algebraical manipulations.

On the other hand by Binomial theorem it is easy to see that $a_{n}$ is increasing if $p > 0$. We will consider this simplified case when $p > 0$. On the contrary assume that $a_{n}$ is not Cauchy. Then there is a $\epsilon > 0$ such that there are infinity of integers $m, n$ with $n > m$ such that $a_{n} - a_{m} \geq \epsilon$. Let us choose the least of integers $m$ and call it $M$. It is then obvious from increasing nature of $a_{n}$ that $a_{n} - a_{M} \geq k\epsilon$ where $k \leq n - M$ and $k \to \infty$ as $n \to \infty$. It follows that $a_{n}$ is unbounded.

However it is easy to see that $a_{n}$ is bounded. We have \begin{align} a_{n} &= 1 + p + \dfrac{1 - \dfrac{1}{n}}{2!}p^{2} + \cdots\notag\\ &\leq 1 + p + \frac{p^{2}}{2!} + \cdots \end{align}

Hence we get a contradiction.

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