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For $n=1,2,3, \dots$, let

$a_n =(1+\frac{1}{n})^n$ and $b_n = (1+\frac{1}{n})^{n+1}$

I have proven that $a_n \leq a_{n+1} \leq b_{n+1} \leq b_n$. Now, deduce that $a_n \leq 3 $ $\forall n$

My approach was to use the binomial theorem on $b_n$ to try and see if something happened. I still think that this is the right approach, but I lack the finesse to see my method through. Can someone help? And bonus points if you can use the binomial theorem, or show that it's silly to use the binomial theorem.

Also, suggestions for tags?

Bob the Builds
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3 Answers3

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$$ (1 + 1/n)^n = \sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} \le \sum_{k=0}^n \frac{n^k}{k!} * \frac{1}{n^k} \le \sum_{k=0}^n \frac{1}{k!} \le 2+ \sum_{k=2}^n \frac{1}{2^{k-1}} < 2+ \sum_{k=2}^{\infty} \frac{1}{2^{k-1}} = 3 $$

For $k \ge 2$ , $\frac{1}{k!}$ will fall faster than $\frac{1}{2^{k}}$ and in the last step we use geometric row

kave
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$a_n\le b_p$ for all $n,p$. In particular $a_n\le b_5=\dfrac{6^6}{5^6}=\dfrac{12^6}{10^6}=\dfrac{2{,}985{,}984}{10^6}$.

Bernard
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$$a_{n}=(1+\frac{1}{n})^n=\\\binom{n}{0}1^n\frac{1}{n}^0+\binom{n}{1}1^{n-1}\frac{1}{n}^1+\binom{n}{2}\frac{1}{n}^2+\binom{n}{3}\frac{1}{n}^3+...\\=1+\frac{n}{n}+\frac{n(n-1)}{2}\frac{1}{n}^2+\frac{n(n-1)(n-3)}{3!}\frac{1}{n}^3+...\\=1+1+\frac{n-1}{2n}+\frac{(n-1)(n-2)}{6n^2}+...\\\leq 1+1+\frac{n}{2n+2}+\frac{(n)(n-1)}{6(n+1)^2}+...=a_{n+1}$$

Khosrotash
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