Prove decay function concaved-up Below decay function were used for solving TVM, for interest rate.
$\displaystyle f_n(x) = \frac{n\;x}{(1+x)^n-1}\qquad$, where $n>1,\;x>-1$
I wanted to show decay function is concaved-up.
In other words, I wanted to show $f_n^{''}(x) > 0$
This is what I get for second derivative, confirmed by CAS.
$\displaystyle f_n^{''}(x) = 
\frac{f_n(x)^2 × (1+x)^n × 2 × (f_n(x) + ({n-1 \over 2})x-1)} {(x+x^2)^2} $
Details not shown, but if we take limit at $x=0$, we have:
$f_n^{''}(0) = \frac{n^2-1}{6} > 0$
For $x≠0$, all we need is to show
$(f_n(x) + ({n-1 \over 2})x-1) > 0$
Or, $f_n(x)$ is above its tangent line, at $x=0$
Is there a simple proof for decay function concaved up ?
Perhaps induction proof, without doing second derivative ?
 A: 
For $x≠0$, all we need is to show
$(f_n(x) + ({n-1 \over 2})x-1) > 0$

Let $R=1+x>0$
$\displaystyle f_n(x) = \frac{n\;(R-1)}{R^n-1} \stackrel{?}{>}\; \left(1 - \left(\frac{n-1}{2}\right)\;(R-1)\right)$
Or, $\displaystyle \; n \stackrel{?}{>} 
\left( \frac{R^n-1}{R-1} \right)
\left(1 - \left(\frac{n-1}{2}\right)\;(R-1)\right)$
We want to locate RHS extremum.
$\displaystyle \frac{d(RHS)}{dR} = 
\frac{1 - ({n^2+n \over 2})\;R^{n-1} + (n^2\!-\!1)\;R^n - ({n^2-n \over 2})\;R^{n+1}} { (R-1)^2}$
From numerator, with $n>1$, we have 3 sign changes, thus 1 or 3 roots.
However, RHS is a polynomial; so does all its derivatives.
Numerator must have 3 roots, with 2 roots cancelled by denominator.
⇒ RHS has only 1 extremum, at the $x=0$ tangent.
All is require is to check curve above, or below tangent line.
For $x → ∞$:
Decay function $f_n(∞) = 0$
Tagent line: $1-\left({n-1 \over 2}\right)(∞) = -∞$
Thus, decay function above tangent line.
For $x≠0:\;(f_n(x) + ({n-1 \over 2})x - 1) > 0\qquad$QED

We can also move everything to one side, and expand polynomial.
$\begin{align}\displaystyle g_n(R) &= 
\left( \frac{R^n-1}{R-1} \right)
\left(1 - \left(\frac{n-1}{2}\right)\;(R-1)\right) - n \\
&= -\left(\frac{n-1}{2}\right)\;R^n + R^{n-1} + R^{n-2} \;+\;...\;+\;R - \left(\frac{n-1}{2}\right)
\end{align}$
$g(1) = -\left(\frac{n-1}{2}\right) + (n-1) -\left(\frac{n-1}{2}\right)= 0$
$g'(1) = -n\left(\frac{n-1}{2}\right) + (n-1)+(n-2)\;+\;...\;+\;1= 0$
$g_n(R)$ have 2 sign changes, 2 repeated roots at R=1
⇒ Only 1 extremum, at the $\;x=0\;$ tangent line.
