Staver's identity relating $\sum_{k=1}^{n}\binom{2k}{k}\frac{1}{k}$ and $\sum_{k=1}^{n}\left(k\binom{n}{k}\right)^{-2}$

I am looking for a proof of the following identity, very relevant for evaluating certain binomial sums $\pmod{p}$:

$$ \sum_{k=1}^{n}\binom{2k}{k}\frac{1}{k} = \frac{2n+1}{3}\binom{2n}{n}\sum_{k=1}^{n}\frac{1}{k^2\binom{n}{k}^2} \tag{1}$$

It is due to Tor B. Staver, Om summasjon av potenser av binomialkoeffisienten, Norsk Mat. Tidsskrift 29 (1947), but I did not manage to get access to the original article. I have attempted a few things, so far unsuccessfully:

A) Exploit $$ \sum_{k=0}^{n}\frac{\lambda^k}{\binom{n}{k}}=(n+1)\sum_{r=0}^{n}\frac{\lambda^{n+1}+\lambda^{n-r}}{(r+1)(1+\lambda)^{n-r+1}}\tag{A}$$ which holds for any $\lambda > -1$, as shown by Bala Sury through Euler's Beta function.

B) Exploit a partial fraction decomposition of the reciprocal squared binomial coefficient, namely $$ \frac{1}{k^2\binom{n}{k}^2} = \sum_{r=0}^{k-1}\frac{\binom{k-1}{r}^2}{(n-r)^2}+2\sum_{r=0}^{k-1}\frac{\binom{k-1}{r}^2}{(n-r)}(H_{k-1-r}-H_{r-1}) \tag{B}$$

C) Exploit the Beta function and symmetry in order to get

$$ \sum_{k=1}^{n}\frac{1}{k^2\binom{n}{k}^2}=\frac{1}{4^n n!^2}\iint_{(-1,1)^2}\frac{(1+x+y+xy)^{n+1}-(1+xy)^{n+1}}{x+y}\,dx\,dy \tag{C}$$

D) Apply Wilf's snake oil method, proving that the LHS and the RHS of (1) have the same OGF. Quite clearly

$$ \sum_{k=1}^{n}\binom{2k}{k}\frac{1}{k} = 2[z^n]\frac{\log(2)-\log(1+\sqrt{1-4z})}{1-z}.\tag{D}$$

E) I am also able to show through the Beta function that $$\begin{eqnarray*} \sum_{r=0}^{n}\frac{1}{\binom{n}{r}^2}&=&2\frac{(n+1)^2}{(n+2)}\sum_{r=0}^{n}\frac{1}{(n-r+1)\binom{n+r+2}{r}}\\&=&2\frac{(n+1)^2}{(n+2)}[z^{n+1}]\left(-\log(1-z)\cdot\phantom{}_2 F_1\left(1,1;n+3;x\right)\right)\\&=& 2(n+1)^2 [z^{2n+3}](1-z)^{n+1}\left(\log(1-z)-\log^2(1-z)\right)\tag{E}\end{eqnarray*}$$

Denote $$F(k,n) = \frac{1}{k^2 \binom{n}{k}^2} \qquad G(k,n) = \binom{2k}{k}\frac{1}{k} \frac{3}{2n+1}\binom{2n}{n}^{-1}$$ $$S(n) = \sum_{k=1}^n F(n,k) \qquad T(n) = \sum_{k=1}^n G(n,k)$$ let's show $S=T$ by showing they satisfy same recurrence (plus some trivial checks on initial conditions).

Recurrence for $T$ is easy. Note that $$(-n-1) G(k,n)+2 (2 n+3) G(k,1+n) = 0$$ summing this over $k$ from $1$ to $n$ gives $$(-n-1) T(n)+2 (2n+3) (T(1+n)-G(n+1,n+1)) = 0$$ that is $$(-n-1)T(n) + 2(2n+3) T(n+1) = \frac{6}{n+1}$$

On the other hand, one checks, with $R(k,n) = \frac{(2 k-3 n-5) (-k+n+1)^2}{(n+1)^2}$, $$\tag{*}(-n-1) F(k,n)+2 (2 n+3) F(k,1+n)=F(k+1,n) R(k+1,n) -F(k,n) R(k,n)$$ note that RHS telescopes in $k$. Summing over $k$ from $1$ to $n-1$, _{$$(-n-1)(S(n)-\frac{1}{n^2}) + 2(2n+3) (S(n+1)-\frac{1}{(n+1)^2} - \frac{1}{n^2(n+1)^2}) = F(n,n)R(n,n)-F(1,n)R(1,n)$$} which rearranges into $$(-n-1)S(n) + 2(2n+3) S(n+1) = \frac{6}{n+1}$$ QED.