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I am trying to derive a general result on moments generating functions and need to prove that, for any $K \ge 1$, the following function:
$$f_K(x) = \frac{\log(1 + x) - \sum_{i = 1}^K \frac{(-1)^{i + 1}}{i} x^i}{(-1)^{K} x^{K + 1}}$$
is decreasing in $\mathbb{R}^+$. Simulations show that this function is indeed decreasing but manipulating its derivative is a mess.. any ideas on how to show this result?
Thank you for your help!
Consider the numerator $$ g(x) = \log(1 + x) - \sum_{i = 1}^K \frac{(-1)^{i + 1}}{i} x^i $$ first. We have $$ g'(x) = \frac{1}{1+x} - \sum_{i = 0}^{k-1} (-x)^i = \frac{1}{1+x} - \frac{1-(-x)^K}{1+x} = \frac{(-1)^K x^K}{1+x} $$ and therefore $$ \tag{*} f_K(x) = \frac{1}{x^{K+1}} \int_0^x \frac{t^K}{1+t} \, dt \, . $$
Inspired by Oliver Díaz' answer we now can substitute $t = x u$ and get $$ \boxed{f_K(x) = \int_0^1 \frac{u^K}{1+xu} \, du} $$ which is clearly decreasing in $x$.
Alternatively we can compute the derivative of $f_K$ in $(*)$: $$ f_K'(x) = \frac{1}{x(1+x)} - \frac{K+1}{x^{K+2}} \int_0^x \frac{t^K}{1+t} \, dt \, . $$ The right-hand side becomes larger if $t$ in the denominator of the integral is replaced by $x$, so that $$ f_K'(x) < \frac{1}{x(1+x)} - \frac{K+1}{x^{K+2}} \int_0^x \frac{t^K}{1+x} \, dt = 0 \, . $$ This proves that $f_K$ is strictly decreasing on $(0, \infty)$.
Here is a slightly different solution based on the integral form of the Taylor approximation of function.
Theorem: Suppose $f$ is a function defined in an open interval $I$ containing a point $a$, and that$f$ has continuous derivative of order $n+1$. Then, for any $x\in I$ $$ f(x) = \sum^n_{k=0}\frac{f^{(k)}(a)}{k!}(x-a)^k+ E_n(x)$$ where $$E_n(x)=\frac{1}{n!}\int^x_a(x-t)^nf^{(n+1)}(t)\,dt$$
This can be shown by induction on $n$; many textbooks on Calculus and real analysis discuss it; also, there are a few postings about it in MSE.
The change of variable $t=t(u)=x+u(a-x)$ yields another expression for $E_n(x)$ which is more useful for the OP's problem:
$$E_n(x)=\frac{(x-a)^{n+1}}{n!}\int^1_0 u^n f^{(n+1)}(x+u(a-x))\,du$$
In the case of the OP, $f(x)=\log(1+x)$ and $a=0$. Notice that the expression of interest is \begin{align} R_K(x):=\frac{E_K(x)}{(-1)^K x^{K+1}}&=\frac{1}{(-1)^K x^{K+1}}\frac{x^{K+1}}{K!}\int^1_0u^K \frac{(-1)^K K!}{(1+x(1-u))^{K+1}}\,du\\ &=\int^1_0\frac{u^K}{(1+x(1-u))^{K+1}}\,du \end{align} From this it is clear that $R_K$ is a decreasing function of $x>0$, since the integrand in the last expression above is positive and a decreasing function in $x$.
If one must, differentiation yields \begin{align} R'_K(x)&=\frac{\partial}{\partial x}\int^1_0\frac{u^K}{(1+x(1-u))^{K+1}}\,du\\ &=\int^1_0\frac{\partial}{\partial x}\frac{u^K}{(1+x(1-u))^{K+1}}\,du\\ &=-(K+1)\int^1_0\frac{u^K(1-u)}{(1+x(1-u))^{K+2}}\,du<0 \end{align}
Edit: to reconcile the reminder obtained by the the theorem above and the one obtained by @MartinR's using first principles, notice the substitution $v=\frac{x-t}{1+t}=\frac{x+1}{1+t}-1$ in $E_n(x)$, equivalently $t=\frac{x+1}{v+1}-1$, yields $dt=-\frac{x+1}{(v+1)^2}dv$ and $$E_n(x)=(-1)^n\int^x_0 v^n\frac{v+1}{x+1}\frac{x+1}{(1+v)^2}\,dv=(-1)^n\int^x_0\frac{v^n}{1+v}\,dv $$
