1. ## nice inequality

$

\begin{array}{l}
prove\;that \\
\\
1 \prec \frac{1}{{1001}} + \frac{1}{{1002}} + \frac{1}{{1003}} + ...... + \frac{1}{{3009}} \prec 1\frac{1}{3} \\
\end{array}

$

2. Originally Posted by fxs12
$

\begin{array}{l}
prove\;that \\
\\
1 \prec \frac{1}{{1001}} + \frac{1}{{1002}} + \frac{1}{{1003}} + ...... + \frac{1}{{3009}} \prec 1\frac{1}{3} \\
\end{array}

$
Euler's constant is

$\displaystyle\lim_{n\rightarrow\infty}\left(\frac{ 1}{1}+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{n}-log_en\right)$

$log_en\approx\left(1+\frac{1}{2}+\frac{1}{3}+....+ \frac{1}{n}\right)-Euler's\ constant$

$\frac{1}{1001}+\frac{1}{1002}+.......+\frac{1}{300 9}\approx\ log_e3009-log_e1000\approx\ log_e3.009$

$e=2.718...$

$e^{\frac{4}{3}}=3.79...$

Therefore

$log_e2.718..

3. thank you , but how didi you get this?

$\log _e n \approx (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ....... + \frac{1}{n}) - Euler's\;cons\tan t$

4. Sorry, that approximation may be a bit on the crude side!!

What you have is the difference between a pair of Harmonic series.

There are alternative quick calculations for that,
and you may be requiring a particular type of solution,
possibly along other lines.

$H_n=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.......... ......+\frac{1}{n}$

One approximation is...

$\frac{1}{2(n+1)}+log_en+\gamma

where $\gamma\approx\ 0.5772156649$ is Euler's constant.

Taking either $\frac{1}{2n}$ or $\frac{1}{2n+1}$

we get approximately $H_{3009}-H_{1000}\approx\ \frac{1}{2(3009)}+log_e3009-\left(\frac{1}{2(1000)}+log_e1000\right)$

or

$\frac{1}{2(3010)}+log_e3009-\left(\frac{1}{2(1001)}+log_e1000\right)$

which is basically approximately the difference between the logarithms