# help for proofs (1 Viewer)

#### mathsbrain

##### Member

I am fine for the first part, not sure how to do the HENCE part...it could have been easily done with letting f(x)=e^x-1-a^2 and showing its always positive...but the HENCE...

#### fan96

##### 617 pages
Where did you get this question?

You probably wouldn't be able to get this result from just $\bg_white e^n \ge {(n+1)^n}/{n!}$ as

$\bg_white \frac{(n+1)^n}{n!} \stackrel{?}{\ge}1+n^2$

is not actually true, even taking $\bg_white n \ge 0$. You'd have to at least start a bit further back.

I doubt this is the intended answer but one hence-y way could be to cook up another integral in a similar fashion that gives you the inequality you want, e.g.

$\bg_white \int_0^a\frac{(1-x)^2}{1+x^2} \, \mathrm dx \ge 0.$

#### CM_Tutor

##### Well-Known Member
I'm pretty sure that the required substitution is $\bg_white e^a = n + 1$ or $\bg_white a = \ln (n + 1)$ in which case the result $\bg_white n! > \frac{(n + 1)^n}{e^n}$ becomes $\bg_white \ln n! > \frac{a}{e}$. We then need to bring in the integral result or some modification (I tried trapezoidal and it doesn't get us there) to get the final result.

I agree with mathbrain, it is certainly a true result as $\bg_white f(x) = e^x - x^2 - 1$ has $\bg_white f'(x) > 0$ for all $\bg_white x \ge 0$ and $\bg_white f(0) = 0$ so $\bg_white f(x) \ge 0$ (and thus $\bg_white e^x \ge x^2 + 1$) for all $\bg_white x \ge 0$.

#### mathsbrain

##### Member
I'm pretty sure that the required substitution is $\bg_white e^a = n + 1$ or $\bg_white a = \ln (n + 1)$ in which case the result $\bg_white n! > \frac{(n + 1)^n}{e^n}$ becomes $\bg_white \ln n! > \frac{a}{e}$. We then need to bring in the integral result or some modification (I tried trapezoidal and it doesn't get us there) to get the final result.

I agree with mathbrain, it is certainly a true result as $\bg_white f(x) = e^x - x^2 - 1$ has $\bg_white f'(x) > 0$ for all $\bg_white x \ge 0$ and $\bg_white f(0) = 0$ so $\bg_white f(x) \ge 0$ (and thus $\bg_white e^x \ge x^2 + 1$) for all $\bg_white x \ge 0$.
damn...this is tough, can anyone help?

#### CM_Tutor

##### Well-Known Member
Extending from my previous post and correcting an error...

\bg_white \begin{align*} e^n &> \frac{(n + 1)^n}{n!} \\ e^n n! &> (n + 1)^n \\ n! &> \frac{(n+ 1)^n}{e^n} \end{align*}

And use the substitution $\bg_white n + 1 = e^a$:

\bg_white \begin{align*} n! &> \frac{(n+ 1)^n}{e^n} \\ &> \frac{(e^a)^n}{e^n} \\ &> e^{a \times n - n} \\ n! &> e^{n(a - 1)} \\ \ln n! &> n(a - 1) \text{ . . . . . . (1)} \end{align*}

Now, returning to the starting inequality, we had:

$\bg_white \ln n! < \displaystyle\int_1^{n+1} \ln x \; dx = (n + 1)\ln (n + 1) - n < \ln (n + 1)!$

Applying the substitution $\bg_white n + 1 = e^a$:

\bg_white \begin{align*} \ln n! &< (n + 1)\ln (n + 1) - n < \ln (n+1)! \\ \ln n! &< e^a \ln e^a - n < \ln [(n+1) \times n!] \\ \ln n! &< ae^a - n < \ln e^a + \ln n! \\ \text{Yielding two results: } \ln n! &< ae^a - n \text{ . . . . . . (2A)} \\ \text{And} \ln n! &> ae^a - n - a \text{ . . . . . . (2B)} \end{align*}

Now, equations (1) and (2A) can be combined to conclude that $\bg_white 0 > -1$... which it true, but not terribly useful:

\bg_white \begin{align*} ae^a - n > \ln n! &> n(a - 1) \\ ae^a - n &> na - n \\ ae^a &> na \\ e^a &> n \\ e^a &> e^a - 1 \\ 0 &> -1 \end{align*}

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 $\bg_white 3 > x$ and that $\bg_white 5 > 3$ then I can say with certainty that $\bg_white 5 > x$. In this case, I start by noting that $\bg_white ae^a > ae^a - n$ allows me to alter {2B):

\bg_white \begin{align*} ae^a &> n(a - 1) = (n + 1 - 1)a \\ &> (e^a - 1)a \\ ae^a &> ae^a - a \\ a &> 0 \end{align*}

Other variants include modifying (1A)'s term $\bg_white n(a - 1)$ like :

\bg_white \begin{align*} ae^a - n&> n(a - 2) \\ ae^a &> na - 2n + n = n(a - 1) \\ &> (e^a - 1)(a - 1) \\ ae^a &> ae^a - e^a - a + 1 \\ e^a &> 1 - a \end{align*}

\bg_white \begin{align*} ae^a &> (n - 1)(a - 1) \\ &> (e^a - 1 - 1)(a - 1) = (e^a - 2)(a - 1) \\ ae^a &> ae^a - 2a - e^a + 2 \\ 0 &> 2 - 2a - e^a \\ e^a &> 2 - 2a \end{align*}

These are getting closer, though the negative coefficient on the $\bg_white a$ is a problem. As an aside, we would be there if we can get $\bg_white e^a > 2a$ as this could be integrated from 0 to $\bg_white a$ to give $\bg_white e^a - e^0 > a^2 - 0^2$, which is the required result.

\bg_white \begin{align*} ae^a - \frac{n}{2} &> (n - 1)(a - 1) \\ 2ae^a &> 2(n - 1)(a - 1) + n = 2an - 2a - 2n + 2 + n = n(2a - 1) - 2a + 2 \\ &> (e^a - 1)(2a - 1) - 2a + 2 \\ 2ae^a &> 2ae^a - 2a - e^a + 1 - 2a + 2 \\ 0 &> 3 - 4a - e^a \\ e^a &> 3 - 4a \end{align*}

\bg_white \begin{align*} ae^a + n &> (n - a)(a - 1) \\ ae^a &> an - a^2 - n + a - n = n(a - 2) - a^2 - a \\ &> = (e^a - 1)(a - 2) - a^2 - a \\ ae^a &> ae^a - a - 2e^a + 2 - a^2 - a \\ 0 &> 2 - 2a - a^2 - 2e^a \\ e^a &> \frac{2 - a - a^2}{2} \end{align*}

I don't see how this approach is going to yield any result $\bg_white e^a > f(a)$ where $\bg_white f(a)$ has positive coefficients on any terms in $\bg_white a$.

I've also been thinking about the geometric interpretations here:
• Consider the function $\bg_white y = \ln x$ plotted from $\bg_white x = 1$ to $\bg_white x = n + 1$ where $\bg_white n \in \mathbb{Z} \ge 2$. Add divisions (for upper and lower rectangles) with width of 1 unit.
• $\bg_white \ln n!$ is the area of the resulting lower rectangles, $\bg_white \ln (n+1)!$ are the upper rectangles, and so the inequality $\bg_white \ln n! < \displaystyle\int_0^{n+1} \ln x \; dx = (n + 1)\ln(n + 1) - n < \ln (n+1)!$
• Taking $\bg_white n + 1 = e^a$ we have a rectangle with one side on the $\bg_white x$-axis covering $\bg_white 0 \leq x \leq n + 1 = e^a$ and one side on the $\bg_white y$-axis covering $\bg_white 0 \leq y \leq \ln (n+1) = a$
• The integral exists entirely within the rectangle, so that the area of the rectangle ($\bg_white ae^a$) is divided into the part under the curve $\bg_white y = \ln x$ (with area $\bg_white ae^a - e^a + 1$) meaning the area above the curve (which is also the area under the curve against the $\bg_white y$-axis has area $\bg_white e^a - 1$
• 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 $\bg_white y$-axis and the diagonal for the large rectangle, and seek that the area against the $\bg_white y$-axis to be greater than $\bg_white a^2$ - but while still using the original result to satisfy the "hence" requirement.