Sum Involving Bernoulli Numbers : $\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}=\frac{2n-1}{2n+1}$ How can we prove that
$$\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}=\frac{2n-1}{2n+1}$$
where $B_{2r}$ are the Bernoulli numbers?
$$\begin{array}{c|c|c|} n & \frac{2n-1}{2n+1} & \sum_{r=1}^n \binom{2n}{2r-1} \frac{B_{2r}}{r} \\ \hline 1 &\frac{1}{3} & \frac{1}{3} \\ 2 &\frac{3}{5} & \frac{3}{5} \\ 3 &\frac{5}{7} &\frac{5}{7} \\ 4 & \frac{7}{9} & \frac{7}{9} \\ 5 & \frac{9}{11} &\frac{9}{11}\end{array}$$
The formula appears to be correct for a lot of values of $n$.
 A: $$
\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}
$$
$$ = 2\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{2r} = \frac{2}{2n+1}\sum_{r=1}^n \frac{(2n+1)(2n)!}{2r(2r-1)!(2n-2r+1)!}B_{2r}
$$
$$
= \frac{2}{2n+1}\sum_{r=1}^n \frac{(2n+1)!}{(2r)!(2n-2r+1)!}B_{2r}
= \frac{2}{2n+1}\left(\sum_{r=0}^n \left(\binom{2n+1}{2r}B_{2r} + \binom{2n+1}{2r-1}B_{2r-1}\right) - \binom{2n+1}{0}B_0-\binom{2n+1}{1}B_1\right)
$$
$$
= \frac{2}{2n+1}\left(\sum_{r=0}^{2n} \binom{2n+1}{r}B_{r} - \binom{2n+1}{0}B_0-\binom{2n+1}{1}B_1\right)
$$
Since,
$$
B_m = -\sum _{k=0}^{m-1}\binom{m}{k}\frac{B_k}{m-k+1} \implies 0 = \sum _{k=0}^{m}\binom{m+1}{k}B_k
$$
So,
$$
 \frac{2}{2n+1}\left(\sum_{r=0}^{2n} \binom{2n+1}{r}B_{r} - \binom{2n+1}{0}B_0-\binom{2n+1}{1}B_1\right)= \frac{2}{2n+1}\left(0-1-(2n+1)(\frac{-1}{2})\right) = \frac{1}{2n+1}\left(2n+1-2 \right) = \frac{2n-1}{2n+1}
$$
A: This identity  also has  a proof using  the technique  of annihilated
coefficient extractors (ACE).
First observe that it is equivalent to
$$\sum_{r=1}^n \frac{2n+1}{2r} {2n\choose 2r-1} B_{2r}
= n - \frac{1}{2}.$$
The left simplifies to
$$\sum_{r=1}^n {2n+1\choose 2r} B_{2r}.$$
Introduce the following generating  function $f(z)$ for this quantity,
which is
$$f(z) = \sum_{n\ge 1} \frac{z^{2n}}{(2n+1)!}
\sum_{r=1}^n {2n+1\choose 2r} B_{2r}.$$
By  the generating  function of  the  Bernoulli numbers  we have  that
$f(z)$ is
$$\sum_{n\ge 1} \frac{z^{2n}}{(2n+1)!}
\sum_{r=1}^n {2n+1\choose 2r} (2r)! [w^{2r}] \frac{w}{e^w-1}.$$
Switch summations to get
$$\sum_{r\ge 1} 
\left( [w^{2r}] \frac{w}{e^w-1} \right)
\sum_{n\ge r} \frac{z^{2n}}{(2n+1-2r)!}$$
which is
$$\sum_{r\ge 1} 
\left( [w^{2r}] \frac{w}{e^w-1} \right)
\sum_{n\ge 0} \frac{z^{2n+2r}}{(2n+1)!}.$$
This in turn simplifies to
$$\sum_{r\ge 1} 
z^{2r} \left( [w^{2r}] \frac{w}{e^w-1} \right)
\sum_{n\ge 0} \frac{z^{2n}}{(2n+1)!}.$$
The first term is the promised annihilated coefficient extractor and
the second is $\sinh(z)/z$ so we get
$$f(z) = \left(-1 + \frac{1}{2} z + \frac{z}{e^z-1}\right)
\frac{\sinh(z)}{z}.$$
We extract coefficients from the three components. First,
$$(2n+1)! [z^{2n}] \left(-\frac{\sinh(z)}{z}\right)
= -(2n+1)! [z^{2n+1}] \sinh(z) = -1.$$
Second, 
$$(2n+1)! [z^{2n}] \left(\frac{1}{2} z \frac{\sinh(z)}{z}\right)
= (2n+1)! [z^{2n}] \frac{1}{2} \sinh(z) = 0.$$
And third,
$$(2n+1)! [z^{2n}] \frac{\sinh(z)}{e^z-1}
= (2n+1)! [z^{2n}] \frac{1}{2}\frac{e^z-e^{-z}}{e^z-1}.$$
This last one needs some rewriting as in
$$\frac{e^z-e^{-z}}{e^z-1}
= 1 + \frac{-e^z + 1 + e^z - e^{-z}}{e^z-1}
\\= 1 + \frac{1 -  e^{-z}}{e^z-1}
= 1 + e^{-z} \frac{e^z -  1}{e^z-1} = 1 + e^{-z}.$$
Therefore the third component is
$$(2n+1)! [z^{2n}] \frac{1}{2} (1 + e^{-z})
\\ = (2n+1)! \times \frac{1}{2} \times \frac{(-1)^{2n}}{(2n)!}
= (2n+1)\times \frac{1}{2} = n + \frac{1}{2}.$$
The sum of the three contributions is
$$n + \frac{1}{2} + (0) + (-1)
= n - \frac{1}{2}$$
precisely as was to be shown.
There is another annihilated coefficient extractor at this
MSE link I and yet another one at this MSE link II. 
