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I am trying to prove the following by mathematical induction: $$\left(1+\frac1{n}\right)^{n}<\left(1+\frac1{n+1}\right)^{n+1}$$ Other proofs without induction are found here: I have to show $(1+\frac1n)^n$ is monotonically increasing sequence. But I am curious whether it can be proved by induction as well.

What I've tried so far:

The original inequality is equivalent to $$(n+1)^{2n+1}<n^n(n+2)^{n+1}$$ So I have to show: $$(n+2)^{2n+3}<(n+1)^{n+1}(n+3)^{n+2}$$ And, $$(n+1)^{n+1}(n+3)^{n+2}=(n+1)\color{red}{n^n}\left(1+\frac1n\right)^n\cdot(n+3)\color{red}{(n+2)^{n+1}}\left(1+\frac1{n+2}\right)^{n+1}$$$$>(n+1)(n+3)\cdot\color{red}{(n+1)^{2n+1}}\left(1+\frac1n\right)^n\left(1+\frac1{n+2}\right)^{n+1}$$$$=(n+1)(n+3)\left(n+2+\frac1n\right)^n\left(n+1+\frac{n+1}{n+2}\right)^{n+1}$$ and I am stuck.

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Kay K.
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  • Well, each proof for a theorem on natural numbers can be made an induction proof artificially by just not using the induction hypothesis, but I guess that's not what you are looking for? – Lukas Betz Jan 08 '17 at 22:56
  • Last expression is not greater than $(n+2)^{2n+3}$ (try couple values of $n$), so you probably need to go back few steps. – Sil Jan 08 '17 at 23:01
  • @LeBtz Right, I was hoping to prove for $n+1$ by using a hypothesis for $n$. But if you know any other type that works for this case, that would be appreciated too. – Kay K. Jan 08 '17 at 23:05
  • @Sil I see. That's what I guessed as $(n+1)(n+3) < (n+2)^2$, for example. So I was thinking that the reverse way may work not the forward way. But I was curious if the simplest induction method ($n$ and then $n+1$) can work or not for this case. – Kay K. Jan 08 '17 at 23:08

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Let us define \begin{align} t_n :=\left(1+\frac{1}{n} \right)^n. \end{align} We shall use the identity \begin{align} t_n =&\ 1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\ldots\\ &\ +\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{n-1}{n}\right). \end{align}

Inductive Step: Assume it holds for $n=k$, i.e. $t_{k-1}\leq t_k$. Then observe \begin{align} t_{k+1}=&\ 1+1+\frac{1}{2!}\left(1-\frac{1}{k+1}\right)+\frac{1}{3!}\left(1-\frac{1}{k+1}\right)\left(1-\frac{2}{k+1}\right)+\ldots\\ &\ +\frac{1}{k!}\left(1-\frac{1}{k+1}\right)\left(1-\frac{2}{k+1}\right)\cdots\left(1-\frac{k-1}{k+1}\right)\\ &\ +\frac{1}{(k+1)!}\left(1-\frac{1}{k+1}\right)\left(1-\frac{2}{k+1}\right)\cdots\left(1-\frac{k}{k+1}\right)\\ \geq&\ 1+1+\frac{1}{2!}\left(1-\frac{1}{k}\right)+\frac{1}{3!}\left(1-\frac{1}{k}\right)\left(1-\frac{2}{k}\right)+\ldots\\ &\ +\frac{1}{k!}\left(1-\frac{1}{k}\right)\left(1-\frac{2}{k}\right)\cdots\left(1-\frac{k-1}{k}\right)=t_k. \end{align} Hence the inductive step holds.

Note: We didn't use the inductive hypothesis.

Jacky Chong
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