Please help proving a sequence is less than a number using induction

Question

I need to prove that $$a_n=\left(1+\frac{1}{n}\right)^n <3$$ using induction. Any help would be great!

Answer 1

$$\begin{align} \left(1+\frac1n\right)^n&=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}\\\\ &=2+\sum_{k=2}^n\binom{n}{k}\frac{1}{n^k}\\\\ &\le 2+\sum_{k=2}^n\frac{1}{k!}\\\\ &\le 2+\sum_{k=2}^\infty \frac{1}{2^{k-1}}\\\\ &=2+\frac{1/2}{1/2}\\\\&=3 \end{align}$$

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Here, we show that $\binom{n}{k}\frac{1}{n^k}\le \frac1{k!}$ for $k\ge 2$. Note that upon expanding $\binom{n}{k}\frac{1}{n^k}$, we obtain

$$\begin{align} \binom{n}{k}\frac{1}{n^k}&=\frac{n!}{k!\,(n-k)!}\frac{1}{n^k}\\\\ &=\frac{1}{k!}\frac{n!}{(n-k)!\,n^k}\\\\ &=\frac{1}{k!}\frac{n(n-1)(n-2)(n-3)\cdots (n-k+3)(n-k+2)(n-k+1)}{n^k}\\\\ &=\frac{1}{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\left(1-\frac3n\right)\cdots \left(1-\frac{k-3}n\right)\left(1-\frac{k-2}n\right)\left(1-\frac{k-1}n\right)\\\\ &\le \frac1{k!} \end{align}$$