Show that log⁡(x+1)−log⁡(x)&lt;1x for x&gt;0

The function log⁡(1+t) is strictly concave and therefore its graph stays under its tangent line at 0: for any t≠0 and t&gt;−1, log⁡(1+t)&lt;t. Your inequality is equivalent to log⁡(x+1)−log⁡(x)=log⁡(1+1x)&lt;1x

Consider f(x)=log⁡(x+1x)=log⁡(1+1x) . Set u=1x and then after making a good figure you will see that log⁡(1+u)≤u and you are done.

Let x&gt;0. f:t→ln⁡(t) is continuous at [x,x+1] and differentiable at (x,x+1), thus by MVT, ∃c∈(x,x+1): f(x+1)−f(x)=(x+1−x)f′(c) ⇒ln⁡(x+1)−ln⁡(x)=1c and x&lt;c&lt;x+1⇔1x+1&lt;1c&lt;1x ⇒1x+1&lt;ln⁡(x+1)−ln⁡(x)&lt;1x

The function log⁡(1+t) is strictly concave and therefore its graph stays under its tangent line at 0: for any t≠0 and t&gt;−1, log⁡(1+t)&lt;t. Your inequality is equivalent to log⁡(x+1)−log⁡(x)=log⁡(1+1x)&lt;1x

Consider f(x)=log⁡(x+1x)=log⁡(1+1x) . Set u=1x and then after making a good figure you will see that log⁡(1+u)≤u and you are done.

Let x&gt;0. f:t→ln⁡(t) is continuous at [x,x+1] and differentiable at (x,x+1), thus by MVT, ∃c∈(x,x+1): f(x+1)−f(x)=(x+1−x)f′(c) ⇒ln⁡(x+1)−ln⁡(x)=1c and x&lt;c&lt;x+1⇔1x+1&lt;1c&lt;1x ⇒1x+1&lt;ln⁡(x+1)−ln⁡(x)&lt;1x

Find out the answer to "Show that log⁡(x+1)−log⁡(x)<1x for x>0"

Cynthia Bell

Answered question

2021-12-31

Show that

\log (x + 1) - \log (x) < \frac{1}{x}

for

x > 0

Answer & Explanation

Shawn Kim

Beginner2022-01-01Added 25 answers

The function

\log (1 + t)

is strictly concave and therefore its graph stays under its tangent line at

0 :

for any

t \neq 0

and

t > - 1

\log (1 + t) < t .

Your inequality is equivalent to

\log (x + 1) - \log (x) = \log (1 + \frac{1}{x}) < \frac{1}{x}

Edward Patten

Beginner2022-01-02Added 38 answers

Consider

f (x) = \log (\frac{x + 1}{x}) = \log (1 + \frac{1}{x})

. Set

u = \frac{1}{x}

and then after making a good figure you will see that

\log (1 + u) \leq u

and you are done.

Vasquez

Expert2022-01-09Added 669 answers

Let $x > 0$ .
$f : t \to \ln (t)$ is continuous at $[x, x + 1]$ and differentiable at $(x, x + 1)$ , thus by MVT,
$\exists c \in (x, x + 1) :$
$f (x + 1) - f (x) = (x + 1 - x) f' (c)$
$\Rightarrow \ln (x + 1) - \ln (x) = \frac{1}{c}$
and
$x < c < x + 1 \Leftrightarrow \frac{1}{x + 1} < \frac{1}{c} < \frac{1}{x}$
$\Rightarrow \frac{1}{x + 1} < \ln (x + 1) - \ln (x) < \frac{1}{x}$

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