Euler sum with Bernoulli numbers

Question

In many sources, I find such equality:
$$\dfrac{1}{n}\sum_{k=1}^n \binom n k B_kB_{n-k}+B_{n-1}=-B_n$$
where $B_1=-\dfrac{1}{2}$
$$$$However, there don't write how to get it. I think that it's necessary to prove such sum through generating functions and convolutions. Unfortunately, no matter how much I derived formulas and converted sums, I can't prove this equality.
$$$$This is my attempt to prove:
$$\dfrac{n!}{n}\sum_{k=0}^n \dfrac{B_{k}}{k!}\dfrac{B_{n-k}}{(n-k)!}=\dfrac{n!}{n}B_n-B_{n-1}-B_n$$
Hence I need to prove next:
$$\sum_{k=0}^n \dfrac{B_{k}}{k!}\dfrac{B_{n-k}}{(n-k)!}=B_n-\dfrac{1}{(n-1)!}(B_{n-1}+B_n)$$
Then I tried to find the generating function for the right and left parts. So I get: $$\sum_{n=0}^\infty(\sum_{k=0}^n \dfrac{B_{k}}{k!}\dfrac{B_{n-k}}{(n-k)!})t^n=\dfrac{t^2}{(e^t-1)^2}$$
So I got a problem.
I can't find $\sum_{n=0}^\infty B_n t^n$. I think this sum is divergent. What I need to do?
$$$$Hope you can help me. Thanks!

Answer 1

Let $f(t) = t/(e^t - 1) = \sum_{n=0}^{\infty} (B_n /n!) t^n$ be the exponential generating function of the Bernoulli numbers. As OP noticed, expanding $f(t)^2$ gives

$$ \frac{t^2}{(e^t - 1)^2} = f(t)^2 = \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} \frac{B_k}{k!} \frac{B_{n-k}}{(n-k)!} \right) t^n. $$

On the other hand, computing $t f'(t)$ gives

$$ \frac{t - t^2}{e^t - 1} - \frac{t^2}{(e^t - 1)^2} = tf'(t) = \sum_{n=1}^{\infty} \frac{B_n}{(n-1)!} t^n. $$

Combining two identities and comparing the coefficient of $t^n$,

$$ \frac{B_n}{n!} - \frac{B_{n-1}}{(n-1)!} - \sum_{k=0}^{n} \frac{B_k}{k!}\frac{B_{n-k}}{(n-k)!} = \frac{B_n}{(n-1)!}. $$

Multiplying $n!$ to both sides and rearranging, we get

$$ -n B_n = n B_{n-1} + \sum_{k=0}^{n} \binom{n}{k} B_k B_{n-k} - B_n. $$

The term $-B_n$ cancels the summand for $k = 0$, and so,

$$ -n B_n = n B_{n-1} + \sum_{k=1}^{n} \binom{n}{k} B_k B_{n-k}. $$

(Notice that the sum now begins with $ k = 1$.) Dividing both sides by $n$ then proves the desired identity.