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I have to prove that $\ln(n+1) \leq \ln(n)+1$ in the Natural Numbers starting from 1.

I tried to use Induction.

Induction base: $\ln(2)\leq \ln(1)+1$ so its ok

Induction step

$\ln(n+2)$ Here i am really struggling to find out how to use my Induction Hypothesis. Can i please get some Hint ?

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  • $\begingroup$ I took the liberty to edit the variable name to $n$ instead of $x$, even if you indicated it is in natural numbers, I think $n$ is a more suitable name. $\endgroup$ – zwim Dec 7 '17 at 1:42
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The statement is equivalent to the following:

$$\ln(n+1) \leq \ln(n)+1 \iff e^{ln(n+1)}\leq e^{\ln(n)+1 }\iff n+1\leq n \cdot e$$

NOTE it is true since $e^x$ is a monotonic increasing function

PROOF BY INDUCTION

Base case: $$n=1 \implies 2\leq e $$

Inductive step:

let's assume true for $n$: $$n+1\leq n \cdot e$$

wewant to prove that: $$n+1\leq n \cdot e \implies (n+1)+1\leq (n+1) \cdot e$$

and it is true:

$$(n+1)+1 \stackrel{\text{inductive hypothesis}}\leq ne+1\leq ne+e=(n+1)\cdot e$$

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  • $\begingroup$ this proof is not valid, there is nothing to justify the last $\le$ $\endgroup$ – zwim Dec 7 '17 at 3:23
  • $\begingroup$ isn't it true by inductive hypothesis? $\endgroup$ – gimusi Dec 7 '17 at 3:29
  • $\begingroup$ ops sorry now I get your point! $\endgroup$ – gimusi Dec 7 '17 at 3:37
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    $\begingroup$ @zwim I've updated the proof by induction. Thanks! $\endgroup$ – gimusi Dec 7 '17 at 4:05
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Hint: $ln(x+1)=ln(x(1+1/x))=ln(x)+ln(1+1/x)$.

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Hint: $\ln(x)+1=\ln(x)+\ln(e)=\ln(ex)$.

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$\ln(n+1)-\ln(n)=\ln(\frac{n+1}{n})=\ln(1+\frac 1n)\le\ln(2)\le 1$

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Let $f(x) = 1 + \log x - \log(1+x)$. Then $f(1) = 1 - \log 2 = \log e - \log 2 > 0$. Since $$ f'(x) = \frac{1}{x} - \frac{1}{1+x} = \frac{1}{x(x+1)} > 0 $$ $f$ is increasing. Therefore, $f(x) > 0$ for all $x \ge 1$ which implies $$ \log(1+x) < 1 + \log x.$$

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