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Why $\frac{1}{n+1}<\log(n+1)-\log(n)<\frac{1}{n}$?

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Because $e>(1+\frac{1}{n})^n$ and $e<(1+\frac{1}{n})^{n+1}$. Those are elementary inequalities: the first comes from the proof $(1+\frac{1}{n})^n$ is strictly increasing; the second from the fact that $(1+\frac{1}{n})^{n+1}$ is strictly decreasing; and both $(1+\frac{1}{n})^n,\ (1+\frac{1}{n})^{n+1}\to e$ as $n\to \infty$. –  Pacciu Mar 27 '11 at 14:34
    
What conditions on n? (for instance, for n=-0.1 this is not true) –  Gottfried Helms Mar 27 '11 at 18:38
    
Just to clarify: my previous comment was addressed to the OP Vafa –  Gottfried Helms Mar 27 '11 at 20:30
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4 Answers

$\log(n+1) -\log(n) = \int_n^{n+1} {1 \over x}\,dx$. The integrand is between ${1 \over n+1}$ and ${1 \over n}$ and the interval of integration is of length $1$, so your integral will be between ${1 \over n + 1}*1 = {1 \over n + 1}$ and ${1 \over n}*1 = {1 \over n}$.

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Or in a different view, use Mean Value Theorem. –  Aryabhata Mar 27 '11 at 17:52
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$\frac{1}{n+1} < \log((n+1)/n)~ \Leftrightarrow ~ 1 < \log((1+1/n)^{n+1})$, which is equivalent to $e < (1+1/n)^{n+1}$. For the second inequality you get $e > (1+1/n)^n$. Both of these inequalities are true: The second one is true, because $e$ is defined as the limit of $(1+1/n)^n$ which is strictly monotonly increasing. The first one is true because $(1+1/n)^{n+1}$ also converges to $e$ and is strictly monotonly decreasing.

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if n is integer>0 then you can do the following:
1a) $ \qquad {1 \over n+1} < \log(1+n) - \log(n) < {1 \over n} $
rewrite:
1b) $ \qquad {1 \over n+1} < \log(1+ {1 \over n} ) < {1 \over n} $

First compare the two rhs terms of 1b). Exponentiation gives:
2) $ \qquad \qquad \qquad \qquad 1+ {1 \over n} < 1 + {1 \over n} + {1 \over n^2 *2!} + \ldots $ which is obvious for $n>0$

Now compare the two lhs terms of 1b). Substitute $m$ for $n+1$ where now $m>1$:
3a) $ \qquad {1 \over m} < \log (1 + {1 \over m-1}) $
exponentiate:
3b) $ \qquad \exp ({1 \over m}) < (1 + {1 \over m-1}) = {m \over m-1} = {1 \over 1 - \frac1m} $
3c) $ \qquad 1 + {1 \over m} + {1 \over m^2 2! } + {1 \over m^3 3! }+\ldots < 1 + {1\over m} +{1\over m^2}+{1\over m^3}+\ldots $ which, by termwise comparision, is obviously true , too.

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Edmund Landau proves

$$1-\frac{1}{x} < \log x < x-1$$

Using the fact that

$$ \log x = \lim_{k\to \infty} k\left(x^{1/k}-1\right)$$

He then states

$$\sum_{v=0}^{k-1} y^v {\leq \choose \geq} k \text{ ; for } {0 < x \leq 1 \choose x \geq 1}$$

Then put for $y>0$

$$y^k-1 = (y-1)\sum_{v=0}^{k-1} y^v \geq k(y-1)$$

Thus if $y = x^{\frac{1}{k}}$

$$x-1 \geq k\left(x^{1/k}-1\right) $$

For the second case

$$\log{\frac{1}{x}} = - \log x$$ thus

$$ \log x = -\log{\frac{1}{x}} \geq 1-\frac{1}{x}$$

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