Inequality $(1+\frac1k)^k \leq 3$
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 $3$
- Is this sufficient?
- What is a better wording than "increasing function"
Solution 1:
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 $$(1+\frac{1}{k})^k\lt e\lt 3$$
Solution 2:
We will use the facts that $ \ln(1+x) < x $ and $ (1 + \frac{1}{k})^k = {\rm e}^{k\,\ln( 1 + \frac{1}{k} )} \,.$
$$ \ln\left( 1 + \frac{1}{k} \right) < \frac{1}{k} \Rightarrow k\,\ln\left( 1 + \frac{1}{k} \right) < 1 \Rightarrow {\rm e}^{k\,\ln( 1 + \frac{1}{k} )} < {\rm e} < 3 $$
$$ \Rightarrow {\left( 1 + \frac{1}{k}\right)}^{\frac{1}{k}} < 3 \,.$$