I came across the following inequality, which should hold for any integer $k\geq 1$:
$$\sum_{j=0}^{k-1}\frac{(-1)^{j}2^{k-1-j}\binom{k}{j}(k-j)}{2k+1-j}\leq \frac{1}{3}.$$
I have been struggling with this statement for a while. It looks valid for small $k$, but a formal proof seems out of reach with my tools. Any suggestions on how to approach this?
For $j=0,\dots,k-1$, \begin{equation*} \frac1{2k+1-j}=\int_0^1 x^{2k-j}\,dx. \end{equation*} So, \begin{equation*} \begin{aligned} s:=&\sum_{j=0}^{k-1}\frac{(-1)^{j}2^{k-1-j}\binom{k}{j}(k-j)}{2k+1-j} \\ &=\int_0^1 dx\,\sum_{j=0}^{k-1}(-1)^{j}2^{k-1-j}\binom{k}{j}(k-j)x^{2k-j} \\ &=\int_0^1 dx\,kx^{k+1}(2x-1)^{k-1}=I_1+I_2, \end{aligned} \tag{1}\label{1} \end{equation*} where \begin{equation*} I_1:=\int_0^{1/2} dx\,kx^{k+1}(2x-1)^{k-1}=(-1)^{k-1}\frac{k!(k+1)!}{(2k+1)!}\le \frac1{2^{2k+2}}\le\frac1{16k^2}, \tag{2}\label{2} \end{equation*} \begin{equation*} I_2:=\int_{1/2}^1 dx\,ke^{g(x)}, \end{equation*} \begin{equation*} g(x):=(k+1)\ln x+(k-1)\ln(2x-1). \end{equation*} Next, $g(1)=0$, $g'(1)=3k-1$, and, for $x\in(1/2,1)$, \begin{equation*} g''(x)=-\frac{k+1}{x^2}-\frac{4(k-1)}{(2x-1)^2}\le-(k+1)-4(k-1)=3-5k \end{equation*} and hence $g(x)\le h(x):= (3k-1)(x-1)+(3-5k)(x-1)^2/2$. So, \begin{equation*} I_2\le\int_{-\infty}^1 dx\,ke^{h(x)}=J(k):=\sqrt{\frac{\pi }{2}} e^{\frac{(1-3 k)^2}{10 k-6}} k\, \frac{\text{erf}\left(\frac{1-3 k}{\sqrt{10 k-6}}\right)+1}{\sqrt{5 k-3}}. \tag{3}\label{3} \end{equation*} Let \begin{equation*} H(k):=\text{erf}\left(\frac{1-3 k}{\sqrt{10 k-6}}\right)+1 -\left(\frac{1}{3}-\frac{1}{16k^2}\right)\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{(1-3 k)^2}{10 k-6}} \sqrt{5 k-3} }{k}. \end{equation*} Then \begin{equation*} H'(k)=\frac{e^{-\frac{(1-3 k)^2}{10 k-6}} \left(160 k^4-647 k^3+75 k^2+456 k-162\right)}{48 \sqrt{2 \pi } k^4 (5 k-3)^{3/2}}>0 \end{equation*} for $k\ge4$ and $H(k)\to0$ as $k\to\infty$. So, for $k\ge4$ we have $H(k)<0$ or, equivalently, \begin{equation*} J(k)<\frac{1}{3}-\frac{1}{16k^2}. \end{equation*} Therefore and in view of \eqref{1}, \eqref{2}, and \eqref{3}, for $k\ge4$ we have \begin{equation*} s<\frac13, \end{equation*} as desired. Checking the latter inequality for $k=1,2,3$ is straightforward.
If the value of the sum is $\frac13-\varDelta(k)$, then it appears that $\varDelta(k)$ satisfies the recurrence $$ (8k+4)\varDelta(k) = (7k-5)\varDelta(k-1) + k\varDelta(k-2).$$ Note that I didn't prove this, though I assume that standard methods for deriving recurrences will suffice. I checked 500 terms. The nonnegativity of $\varDelta(k)$ of course follows trivially from the recurrence and the first few values.
