Extending from my previous post and correcting an error...
We start with the proven result:
And use the substitution
![](https://latex.codecogs.com/png.latex?\bg_white n + 1 = e^a)
:
Now, returning to the starting inequality, we had:
Applying the substitution
![](https://latex.codecogs.com/png.latex?\bg_white n + 1 = e^a)
:
Now, equations (1) and (2A) can be combined to conclude that
![](https://latex.codecogs.com/png.latex?\bg_white 0 > -1)
... which it true, but not terribly useful:
The approach can be modified, however, increasing or decreasing one side of an inequality where the change cannot falsify the statement. This is akin to saying that if
![](https://latex.codecogs.com/png.latex?\bg_white 3 > x)
and that
![](https://latex.codecogs.com/png.latex?\bg_white 5 > 3)
then I can say with certainty that
![](https://latex.codecogs.com/png.latex?\bg_white 5 > x)
. In this case, I start by noting that
![](https://latex.codecogs.com/png.latex?\bg_white ae^a > ae^a - n)
allows me to alter {2B):
Other variants include modifying (1A)'s term
![](https://latex.codecogs.com/png.latex?\bg_white n(a - 1))
like :
These are getting closer, though the negative coefficient on the
![](https://latex.codecogs.com/png.latex?\bg_white a)
is a problem. As an aside, we would be there if we can get
![](https://latex.codecogs.com/png.latex?\bg_white e^a > 2a)
as this could be integrated from 0 to
![](https://latex.codecogs.com/png.latex?\bg_white a)
to give
![](https://latex.codecogs.com/png.latex?\bg_white e^a - e^0 > a^2 - 0^2)
, which is the required result.
I don't see how this approach is going to yield any result
![](https://latex.codecogs.com/png.latex?\bg_white e^a > f(a))
where
![](https://latex.codecogs.com/png.latex?\bg_white f(a))
has positive coefficients on any terms in
![](https://latex.codecogs.com/png.latex?\bg_white a)
.
I've also been thinking about the geometric interpretations here:
- Consider the function
plotted from
to
where
. Add divisions (for upper and lower rectangles) with width of 1 unit.
is the area of the resulting lower rectangles,
are the upper rectangles, and so the inequality ![](https://latex.codecogs.com/png.latex?\bg_white \ln n! < \displaystyle\int_0^{n+1} \ln x \; dx = (n + 1)\ln(n + 1) - n < \ln (n+1)!)
- Taking
we have a rectangle with one side on the
-axis covering
and one side on the
-axis covering ![](https://latex.codecogs.com/png.latex?\bg_white 0 \leq y \leq \ln (n+1) = a)
- The integral exists entirely within the rectangle, so that the area of the rectangle (
) is divided into the part under the curve
(with area
) meaning the area above the curve (which is also the area under the curve against the
-axis has area ![](https://latex.codecogs.com/png.latex?\bg_white e^a - 1)
- This provides insight into some of the inequalities and the reason for the substitution, but does not yield a solution.
- I do wonder, though, if we need to work against the
-axis and the diagonal for the large rectangle, and seek that the area against the
-axis to be greater than
- but while still using the original result to satisfy the "hence" requirement.