Problem: for x>0

    1/2 - (1/8)*x  <  1/x - 1/(e^x - 1)  <  1/2

>The inequality above is in the middle of a proof in
>
>S. K. Lakshmana Rao, "On the sequence for Euler's constant",
>Amer. Math. Monthly 63 (1956), 572-573.
>
>This inequality is supposed to follow from tanh(x) < x
>(and I can verify this as well), but I'm not getting anywhere.

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Here's a hint which I'm pretty sure will work.

Recall that (d/dx) tanh(x)  =  1 - tanh(x)^2  > 1 - x^2

so by integrating, we have  tanh(x) > x - x^3/3

and go from there.

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The left inequality isn't too hard if you proceed in a naive way.

Rearranging, it's equivalent to
e^x > (x^2 + 4x + 8)/(x^2 - 4x + 8).
Let 
f(x) = (x^2 - 4x + 8)e^x - (x^2 + 4x + 8).
Then
f'(x) = (x^2 - 2x + 4)e^x - (2x + 4)
and
f''(x) = (x^2 + 2)e^x - 2.
Then f(0) = f'(0) = 0 and for x > 0,
f''(x) = x^2 e^x + 2(e^x -1) > 0.
This gives in turn, f'(x) > 0 and f(x) > 0 for x > 0,
and the last inequality is the one we want.

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>this looked so nice, that I tried my hand and found
>this lovely expression:
>
>    e^x  >  ( x^2 + 4x + 8) / ( x^2 - 4x + 8)
>
>which is equivalent to your inequality.

   With a little more work, you can translate this to

     tanh(x/2)  >  4*x/(x^2 + 8)            (x > 0)

You can check these by letting

     f(x) = x - 2 arctanh( 4*x/(x^2 + 8))
          = x + ln(x^2 - 4*x + 8) - ln(x^2 + 4*x + 8).

Obviously f(0) = 0.  Whichever form you choose
to differentiate, f'(x) is a rational function in x
(this is why we are using logarithms).  A calculation gives

       f'(x) = x^2 (x^2 + 8) / ((x^2 - 4*x + 8) (x^2 + 4*x + 8)),

which is non-nonnegative for x >= 0.

    Looking at the power series for tanh(x/2), we may guess

           tanh(x/2) > 6*x/(x^2 + 12)        (x > 0)

This is easily checked in the same way.
Going backwards, we can replace the constant
1/8 by 1/12 in the original problem.  Setting x = 1 in

         1/2 - x/12 < 1/x - 1/(e^x - 1)

gives  0.41667 < 0.41802, much tighter that 0.37500 < 0.41802.

>I am not sure whether this will help. But aren't there
>technics to do the polynom division and compare
>the outcome to the power series expansion of e^x ?

-- 
The 21st century is starting after 20 centuries complete,
but we say someone is age 21 after 21 years (plus fetus-hood) complete.
        Peter-Lawrence.Montgomery@cwi.nl    Home: San Rafael, California
        Microsoft Research and CWI

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The secret here seems to be that  z/(e^z - 1)  is the series for
Bernoulli
numbers, which (after some adjustment to the first couple of terms)
is one of those series with alternating signs where truncating it at any

order gives valid upper and lower bounds (for the function at all x>0,
not just where the series converges).  I think the right adjustment
of the function where this is literally true was z*coth(z) (consistent
with the remark below), but I don't remember precisely.