Proving the bound $\left ( 1+\frac{x}{n} \right)^n \leqslant 3^x$, $\forall x \in \mathbb{R^+}$

Question

I'm trying to directly prove the above bound. I have tried expanding it

$$\left ( 1+\frac{x}{n} \right)^n = \sum_{k\geqslant 0} \binom{n}{k}\left ( \frac{x}{n}\right)^k$$

$$= \sum_{k=0\dots n }x^k\cdot \frac{n\cdot(n-1)\cdots(n-k+1)}{n^k}\cdot \frac{1}{k!}$$

$$\leqslant \sum_{k=0\dots n} \frac{x^k}{k!} \leqslant \sum_{k=2\dots n} \frac{x^k}{k(k-1)}=\sum_{k=2\dots n} \frac{-x^k}{k}+\sum_{k=2\dots n} \frac{x^k}{k-1}$$

but can't proceed from here.

I'm not sure it's the right strategy.

Thanks

Answer 1

Hint: Since $2^{k-1}=(2)(2)\cdots (2)<(2)(3)\cdots (k)=k!$ for $k\geq 2$, we have $$\left(1+{1\over n}\right)^n=\sum_{k=0}^n{n\choose k}\left({1\over n}\right)^k\leq \sum_{k=0}^\infty{1\over k!}\leq 1+1+\sum_{k=2}^\infty {1\over 2^{k-1}} =3.$$

---

Hint 2: If $x,n$ are positive integers, then $(1+{x\over n})\leq (1+{1\over n})^x$. Raising this to the $n$th power, then using the bound above gives $$\left(1+{x\over n}\right)^n\leq \left(1+{1\over n}\right)^{nx}\leq 3^x$$

Answer 2

If we can use the limit definition of $e$, then we have $\lim_{n\to\infty } \left(\left(1+\frac{x}{n}\right)^{n/x}\right)^x = e^x < 3^x$.