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How can I elegantly show that: $(1 + \frac{1}{k})^k \leq 3$ For instance I could use the fact that this is an increasing function and then take $\lim_{ k\to \infty}$ and say that it equals $e$ and therefore is always less than $3$

  1. Is this sufficient?
  2. What is a better wording than "increasing function"
Martin Sleziak
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CodeKingPlusPlus
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  • A bounded monotonic sequence ${x_n}$ is convergent. – Mikasa Sep 21 '12 at 06:16
  • 1.- You only have to show that the function/sequence is monotonic increasing ($a_{n+1}>a_{n} \forall n \in \mathbb{N}$). And you can use the value of the limit if known to compare it to the given value. – 3d0 Jun 22 '15 at 14:37
  • The algebraic expression $\bigg(1 + \dfrac 1k\bigg)^k$ is never equal to $3$, but is always less than $3$. $$\lim_{k\to\infty}\bigg(1+\frac 1k\bigg)^k = e = 2.7182818284590452353602874713526\ldots < 3$$ – Mr Pie Feb 18 '18 at 04:21

2 Answers2

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Using binomial theorem, $(1+\frac{1}{k})^k=1+{k\choose 1}\frac{1}{k}+{k\choose 2}\frac{1}{k^2}+\cdots+{k\choose r}\frac{1}{k^r}+\cdots+{k\choose k}\frac{1}{k^k}$

Now, consider ${k\choose r}\frac{1}{k^r}=\frac{k!}{r!(k-r)!}\frac{1}{k^r}=\frac{(1)(1-\frac{1}{k})(1-\frac{2}{k})...(1-\frac{r-1}{k})}{r!}\lt \frac{1}{r!}$

Thus, $$(1+\frac{1}{k})^k=1+{k\choose 1}\frac{1}{k}+{k\choose 2}\frac{1}{k^2}+\cdots+{k\choose r}\frac{1}{k^r}+\cdots+{k\choose k}\frac{1}{k^k}$$ $$ \lt 1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{r!}+\cdots+\frac{1}{k!}$$ $$\lt 1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{r!}+\cdots+\frac{1}{k!}+\frac{1}{(k+1)!}+\cdots =e$$

Hence $$(1+\frac{1}{k})^k\lt e\lt 3$$

Aang
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We will use the facts that $ \ln(1+x) < x $ and $ (1 + \frac{1}{k})^k = {\rm e}^{k\,\ln( 1 + \frac{1}{k} )} \,.$

$$ \ln\left( 1 + \frac{1}{k} \right) < \frac{1}{k} \Rightarrow k\,\ln\left( 1 + \frac{1}{k} \right) < 1 \Rightarrow {\rm e}^{k\,\ln( 1 + \frac{1}{k} )} < {\rm e} < 3 $$

$$ \Rightarrow {\left( 1 + \frac{1}{k}\right)}^{\frac{1}{k}} < 3 \,.$$

Mhenni Benghorbal
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