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Prove with induction on $n$ that \begin{align*} \Bigl(1+ \frac{1}{n}\Bigr)^n < n \end{align*} for natural numbers $n \geq 3$.

Attempt at proof: Basic step. This can be verified easily.

Induction step. Suppose the assertion holds for $n >3$, then we now prove it for $n+1$. We want to prove that \begin{align*} \big( 1+ \frac{1}{n+1})^{n+1} < n+1. \end{align*} So we have \begin{align*} \big( 1+ \frac{1}{n+1}\big)^{n+1} &= \big( 1+ \frac{1}{n+1} \big)^n \cdot \big( 1 + \frac{1}{n+1} \big) \\ & < \big (1 + \frac{1}{n} \big)^n \cdot \big( 1 + \frac{1}{n+1} \big) \\ & = n \cdot \big( 1 + \frac{1}{n+1} \big) \qquad \text{(Induction hypothesis)} \\ &= n \cdot \big( \frac{n+1+1}{n+1} \big) \\ &= \frac{n^2 + 2n}{n+1} \\ &= \frac{(n+1)^2 -1}{(n+1)} \end{align*} And now I'm stuck. I don't know how to get $n+1$ on the RHS. Please help!

Przemysław Scherwentke
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Kamil
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3 Answers3

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HINT: $$ \frac{(n+1)^2-1}{n+1}<\frac{(n+1)^2}{n+1}. $$

Przemysław Scherwentke
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For $n=3$ we have $(1 + 1/n)^{n} < n$. If $n \geq 1$ such that $(1 + 1/n)^{n} < n$, then $$ (1 + \frac{1}{n+1})^{n+1} < (1 + \frac{1}{n})^{n+1} < n(1 + \frac{1}{n}) = n+1, $$ qed.

Yes
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First: When using the induction hypothesis it should read as $$\left(1+\frac{1}{n}\right)^n\cdot\left(1+\frac{1}{n+1}\right)\color{red}<\color{black}n\cdot\left(1+\frac{1}{n+1}\right)$$

Then: $$(n+1)^2-1<(n+1)^2,$$ so we get...can you take it from here?

Hirshy
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