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How to show the sequence $x_n = (1 + \frac{x}{n})^{n}$ is bounded above by $e^x$?

Note: I'm not supposed to be able to use any differentiation techniques if possible. Since we techincally "don't know" it yet.

As can be deduced I am trying to show that the sequence $x_n = (1 + \frac{x}{n})^{n}$ is convergent. I have to arrive at this conclusion using the monotonic convergence theorem. So we are given by definition that $$e^{x} = \lim_{n \rightarrow \infty} \Bigg(1 + \frac{x}{n} \Bigg)^{n}$$

I think I figured out how to show the sequence is montonically increasing. My problem is showing that it is bounded. So one idea I thought of trying to apply was using the binomial theorem:

$$\Bigg(1 + \frac{x}{n}\Bigg)^{n} = \sum_{k = 0}^{n} \binom{n}{k} \bigg(\frac{x}{n}\Bigg)^{k}$$ and then since this would be some sort of finite quantity, I would compare it to $e^x$:

$$\Bigg(1 + \frac{x}{n}\Bigg)^{n} = \sum_{k = 0}^{n} \binom{n}{k} \bigg(\frac{x}{n}\Bigg)^{k} < e^{x} = \lim_{n \rightarrow \infty} \Bigg(1 + \frac{x}{n} \Bigg)^{n} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots $$

But I can't seem to get the finite binomial expansion in a comparable form.

Questions:

1) Is this a correct approach ?

2) If it is how can I rewrite the binomial expansion to work in my favor ?

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  • $\begingroup$ $e^x$ eventually exceeds any polynomial, as can be shown with repeated applications of L’hopital’s rule $\endgroup$ – J. W. Tanner May 23 at 0:10
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    $\begingroup$ If it's monotonically increasing and it has a limit, doesn't that tell us that it's bounded by the limit? $\endgroup$ – Jack Crawford May 23 at 0:10
  • $\begingroup$ @JackCrawford: given the context of the problem your comment is all one needs to write as an answer. I don't see any need for exponential series etc. +1 for your comment. $\endgroup$ – Paramanand Singh May 27 at 15:38
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If $x>0$, how $$\Bigg(1 + \frac{x}{n}\Bigg)^{n} = \sum_{i = 0}^{n} \binom{n}{i} \bigg(\frac{x}{n}\bigg)^{i}$$ then if $i=k$, the $k-th$ term in this serie is:

$$ \binom{n}{k} \bigg(\frac{x}{n}\bigg)^{k}=\frac{x^k}{k!}\cdot\frac{n!}{(n-k)!n^k}$$ Note that $$\frac{n!}{(n-k)!n^k}\leq1$$ Therefore $$ \binom{n}{k} \bigg(\frac{x}{n}\bigg)^{k}=\frac{x^k}{k!}\cdot\frac{n!}{(n-k)!n^k}\leq \frac{x^k}{k!}$$

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  • $\begingroup$ To use this result I would have to establish that $\frac{n!}{(n-k)!n^k}\leq1$ is true. Would I do that by induction? $\endgroup$ – dc3rd May 23 at 1:32
  • $\begingroup$ Yes a way is induction. $\endgroup$ – AsdrubalBeltran May 23 at 1:34
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The result is not true for $x<0$. A better approach for $x >0$: If $x >0$ the $(1+\frac x n)^{n}=e^{n\log (1+x/n)}\leq e^{n\frac x n}=e^{x}$ where I have used the inequality $\log(1+y) \leq y$ for all $y >0$.

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Hint: If $t>0$ then $$\log(1+t)=\int_1^{1+t}\frac{ds}s<\int_1^{1+t}ds=t.$$

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For $x\ge 0 $, $(1+\frac{x}{n})^n=1+x+\frac{n(n-1)x^2}{2n^2}+\frac{n(n-1)(n-2)x^3}{6n^3}+...\lt 1+x+\frac{x^2}{2}+\frac{x^3}{6}+...=e^x$ Therefore the term is bounded and increasing with $n$

Essentially you have $\frac{\binom{n}{k}}{n^k}\le \frac{1}{k!}$

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  • $\begingroup$ For me to use this result I have to first establish that: $\frac{\binom{n}{k}}{n^k}\le \frac{1}{k!}$ is true. Would I be able to establish this through induction? $\endgroup$ – dc3rd May 23 at 0:37
  • $\begingroup$ I am not sure. The observation is simple enough. $\frac{n(n-1)....(n-k+1)}{n^k}\le 1$. $\endgroup$ – herb steinberg May 23 at 0:43
  • $\begingroup$ @dc3rd Also if you look at what happens for fixed $k$ as $n\to \infty$, $\frac{n(n-1)...(n-k+1)}{n^k}$ increases $\to 1$. $\endgroup$ – herb steinberg May 23 at 2:35
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With calculus and elementary method, that is by MVT for $y=\ln(1+t)$ $$\ln(1+a)-ln(a)=\dfrac{1}{\xi}$$ where $a<\xi<a+1$ which shows $$\ln(1+\dfrac1a)=\dfrac{1}{\xi}<\dfrac{1}{a}$$ or $\left(1+\dfrac1a\right)^a<e$ now let $a=\dfrac{n}{x}$.

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From your question it appears that the goal here is to justify that the definition $$e^x=\lim_{x\to\infty} \left(1+\frac{x}{n}\right)^n\tag{1}$$ makes sense. Otherwise if one assumes the existence of the above limit the boundedness of the corresponding sequence is automatically guaranteed (the comment by Jack Crawford and my response to it are based on this assumption).

I hope you are familiar with the proof (given in most textbooks) that the sequence $(1+(1/n))^n$ is increasing and bounded above (say by $3$) and hence it converges to a limit conventionally denoted by $e$. If $k$ is a positive integer then we have $$\left(1+\frac{k}{n}\right)^n=\prod_{i=1}^{k}\left(1+\frac{1}{n+i-1}\right)^{n+i-1}\cdot\left(1+\frac{1}{n+i-1}\right)^{1-i}\tag{2}$$ Each term in the product on right side tends to $e\cdot 1=e$ and hence the overall product tends to $e^k$.

What we have proved here is that the sequence in question is convergent if $x$ is a positive integer. To deal with the case when $x$ is positive but not necessarily an integer we need to take a positive integer, say $k$, such that $k>x$ (we may take $k=\lfloor x\rfloor +1$). Then we have $$\left(1+\frac{x}{n}\right)^n\leq \left(1+\frac {k} {n} \right) ^n\tag{3}$$ The right side above converges to $e^k$ and hence is abounded above. Consequently the left side is also bounded above.

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