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I was asked by a junior of mine to explain to her how the following is true:

For all $x\geq 1$, \begin{eqnarray} 2\leq \left(1+\frac{1}{x}\right)^{x}<2.8. \end{eqnarray}

I know that $\left(1+\frac{1}{x}\right)^{x}\to \text{e}\approx2.7$ as $x\to \infty$. However, she does not have the requisite background to understand limits.

Can someone point me out to a more preliminary level at which this problem can be dealt?

Mark Viola
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P. N. Karthik
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    The left inequality is pretty simple: By the binomial theorem $$\big (1 + \frac{1}{x} \big)^x \geq 1 + x \cdot \frac{1}{x} = 2$$ with inequality iff $x = 1$ – Airdish Nov 12 '16 at 18:27
  • When you say "for all $x$", do you mean for all natural numbers $x$ or all real numbers $x$? Because if it's the real case then things are a bit more complicated and even the previous comment doesn't work (there is no binomial theorem for real exponents, at least not at this level). – Arthur Nov 12 '16 at 18:32
  • Oops I meant "equality" in my last comment, not "inequality", and yeah the distinction between natural numbers and real numbers is very important. – Airdish Nov 12 '16 at 18:34
  • @Arthur The question asks us to show the result for all $x\geq 1$, $x\in \mathbb{R}$. – P. N. Karthik Nov 12 '16 at 18:38
  • What can we use ? Logarithms ? Derivations ? Some calculus will be needed to prove this inequality. – Peter Nov 12 '16 at 18:43
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    Yes. Even I guess so. I was trying to think of the simplest possible way in which I could explain this to her. But I think we may need to use derivatives. – P. N. Karthik Nov 12 '16 at 18:45
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    But a derivative is a limit and we can't use limits. I don't see how we can prove that $f(x)=\left(1+\frac{1}{x}\right)^x$ is an increasing function on $[1,+\infty)$ without derivative. – Michael Rozenberg Nov 12 '16 at 18:46
  • That is very true. She does not possess any knowledge of limits. Thus, I am trying to look for a solution that doesn't use limits anywhere. But I do not know if it can be circumvented.

    Can this result be shown for positive integers without the use of limits? Is it through binomial theorem?

     – P. N. Karthik Nov 12 '16 at 18:48
  • Yes, of course, but it's another problem. Change a given and you'll get a proof. – Michael Rozenberg Nov 12 '16 at 18:50

4 Answers4

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I thought it might be useful to present a way forward in which $x$ is any positive integer. To that end we proceed.

In THIS ANSWER, I showed using only Bernoulli's Inequality that $\left(1+\frac1n \right)^n$ is monotonically increasing for integer values of $n\ge 1$.

From the binomial theorem, we have for $n\ge 1$

$$\begin{align} \left(1+\frac1n \right)^n&=1+1+\frac1{2!} \left(1-\frac1n\right)+\frac1{3!}\left(1-\frac1n\right)\left(1-\frac2{n}\right)+\cdots \frac1{n!}\left(1-\frac{n-1}{n}\right)\\\\ &\le \sum_{k=0}^n\frac1{k!}\\\\ &\le \sum_{k=0}^\infty\frac1{k!}\\\\ &= 1+1+\frac12+\frac16+\sum_{k=4}^\infty \frac{1}{k!}\\\\ &\le 2\frac23+\sum_{k=4}^\infty \frac{1}{2^k}\\\\ &=2\frac23+\frac18\\\\ &=2\frac{19}{24}\\\\ &<2.8 \end{align}$$

From the monotonicity, the lower bound is evidently $2$ (i.e., the term of interest is greater than its value at $n=1$). Putting it together reveals

$$2\le \left(1+\frac1n\right)^n<2.8$$

for integer values of $n\ge 1$.

Community
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Mark Viola
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Let $f(x)=(1+\frac1x)^x$ for $x\geq 1$.

$g(x)=\ln(f(x))=x\ln(\frac{1+x}{x})$

$g'(x)=\ln(1+x)-\ln(x)-\frac{1}{1+x}$

$=\frac{1}{c_x}-\frac{1}{1+x}\;\;$ by MVT.

$>0\;\;$ since $\;\;x<c_x<1+x$.

thus

$g\;$ and $\;f$ are increasing and

$\forall x\geq 1$

$$f(1)=2<f(x)<\lim_{x\to+\infty}f(x)=e<2.8$$

hamam_Abdallah
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This is partial answer since you need to be able to show that given function is increasing/decreasing, for which you might need to use derivations (but for integer numbers not necessarily so).

So assuming you are able to show that $\left(1+\frac{1}{x}\right)^x$ is increasing for $x \geq 1$, lower bounds follows, since you have

$$2 = \left(1+\frac{1}{1}\right)^1 \leq \left(1+\frac{1}{x}\right)^x.$$

On the other hand if you are able to show that $\left(1+\frac{1}{x}\right)^{x+\frac{1}{2}}$ is decreasing for $x\geq 1$, the upper bound follows for $x \geq 2$ since

$$\left(1+\frac{1}{x}\right)^x \leq \left(1+\frac{1}{x}\right)^{x+\frac{1}{2}} \leq \left(1+\frac{1}{2}\right)^{2+\frac{1}{2}} \approx 2.775$$

Sil
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Let $h(x)=\ln(1+x)-x\;\;$ for $x\geq0$.

$$h'(x)=\frac{1}{1+x}-1$$

$$=\frac{ -x}{ 1+x }\leq 0$$

thus

$$(\forall x>0)\; \;\; h(x)\leq h(0)$$

$\implies$

$$(\forall x>0)\;\;\ln(1+x)\leq x$$

or

$$(\forall x>0)\;\; \ln(1+\frac 1x)\leq \frac 1x$$

$\implies$

$$(\forall x>0)\;\; x\ln(1+\frac 1x)\leq \ln(e)$$

And finally

$$(\forall x\geq 1)\;\; 2\leq (1+\frac 1x)^x\leq e$$

hamam_Abdallah
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