An inequality involving binomial coefficients and the powers of two 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?
 A: 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.
A: 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.
