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Here's the problem:

Use the principle of mathematical induction to prove that $n \lt (\frac{3}{2})^n$ for all integers $n \ge 1$.

Here's what I've got:

Base Case: $1 \lt (\frac{3}{2})^1$ is true.

Induction Step: I want to show that $n+1 \lt (\frac{3}{2})^{n+1}$. Doing some algebra gets me to

$$ n<\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)^n - 1 $$

which, sure, is obviously true; it does, however, not a proof make. I know that I'm doing the induction step, in some sense, "backwards," but I don't really know how else to go about it. It looked to me like something I could reduce to Bernoulli's inequality, but I couldn't get there. Anyway, any and all help is appreciated. Thanks.

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Suppose $$n\lt \left(\frac32\right)^n$$ then $$\left(\frac32\right)^{n+1}=\left(\frac32\right)\left(\frac32\right)^n\gt\frac32n\gt n+1\,\,\,\,\,\forall n\ge 3$$ Therefore verify your assertion for $n=1, 2$ separately.
Then Mathematical Induction will leads to your answer.

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