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I am trying to prove by induction that $\sqrt[n]{n}<2-\frac{1}{n}$ where $n\ge2$. It seemed simple at first, but I am stuck with $log(2n-1)$ in the RHS. I am in an elementary undergraduate Maths course. Please help me out.

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  • $\begingroup$ This is my first ever question on this platform. Help me improve. $\endgroup$ – Saswat Pati Jun 7 at 7:23
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    $\begingroup$ There is a simple (non-induction) proof in this question: math.stackexchange.com/q/1322898. $\endgroup$ – Martin R Jun 7 at 7:31
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You need to show , $$n\leq \left(2-\frac{1}{n}\right)^n$$ It holds for the base case $n=2$.

Assuming it holds for $n=k$,

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

we need to show the same for $n=k+1$ i.e.,

$$k+1<\left(2-\frac{1}{k+1}\right)^{k+1}$$

let's try to prove something stronger,

$$k+1<\left(2-\frac{1}{k}\right)^{k+1}<\left(2-\frac{1}{k+1}\right)^{k+1}$$

using the the relation for $n=k$,

$$\left(2-\frac{1}{k}\right)^{k+1}>\left(2-\frac{1}{k}\right)\times k=2k-1>k+1 \qquad\forall k>2$$

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By Bernoulli $$\left(2-\frac{1}{n}\right)^n=\left(1+\left(1-\frac{1}{n}\right)\right)^n>1+n\left(1-\frac{1}{n}\right)=n.$$

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    $\begingroup$ “I am trying to prove by induction ...” $\endgroup$ – Martin R Jun 7 at 7:40

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