Sum a product of Bernoulli numbers and binomial coefficients Context: I am interested in developing the large-$x$ asymptotic series of the digamma function
$$\psi\Big(\frac{1}{2}+ix\Big)$$
for real positive $x$.
For this I am using the known asymptotic expansion of $\psi(x)$:
$$\psi(z) \sim \ln z - \frac{1}{2z} - \sum_{n=1}^\infty \frac{B_{2n}}{2n z^{2n}}, \qquad z\rightarrow \infty$$
where $B_{2n}$ are the Bernoulli numbers, in conjunction with the generalized binomial expansion for negative powers:
$$\frac{1}{(1+y)^p} = \sum_{n=0}^\infty \begin{pmatrix}-p\\n\end{pmatrix} y^n$$.
Question:  In the process of applying the binomial expansions to the third term of the large $x$ behavior of the digamma function, I end up with the following double sum:
$$S = -\sum_{d=2}^\infty \Big(\sum_{n=1}^{\lfloor d/2 \rfloor} \frac{B_{2n}}{2n} \begin{pmatrix}-2n\\d-2n\end{pmatrix}2^{2n}\Big)\big(\frac{-i}{2x}\big)^d$$
What is the closed-form formula for the inner $n$-sum, for $d$ odd and for $d$ even?  I believe this should be possible since the odd $d$ case leads to a series that is purely imaginary which (I observe by explicit computation of the first few terms) exactly cancels the imaginary part arising from the expansion of $\ln(z) - \frac{1}{2z}$.  I am therefore able to guess
$$-\sum_{n=1}^{\lfloor d/2 \rfloor} \frac{B_{2n}}{2n} \begin{pmatrix}-2n\\d-2n\end{pmatrix}2^{2n} = 1-\frac{1}{d},\qquad d = \{3,5,7,\ldots\}$$

*

*How can I prove this relation? And what is the closed-form result for $d$ even and positive?

*Bonus: what is the large $x$ behavior of $\psi(\frac{1}{2}+ix)$?


We seek to evaluate
$$\sum_{q=1}^{\lfloor n/2\rfloor} \frac{B_{2q}}{2q}
{-2q\choose n-2q} 2^{2q}.$$
Taking advantage of the odd index Bernoulli numbers being zero except
$B_1$ this is
$$(-1)^{n+1} + \sum_{q=1}^n \frac{B_{q}}{q}
{-q\choose n-q} 2^q.$$
Now with the binomial coefficient we get
$$(-q)^{\underline{n-q}}/(n-q)! =
(-1)^{n-q} (n-1)^{\underline{n-q}}/(n-q)! =
(-1)^{n-q} {n-1\choose n-q}.$$
Continuing,
$$(-1)^{n+1} +
\sum_{q=1}^n \frac{B_q}{q} (-1)^{n-q} {n-1\choose q-1} 2^q
= (-1)^{n+1} +
\frac{1}{n} \sum_{q=1}^n B_q (-1)^{n-q} {n\choose q} 2^q
\\ = (-1)^{n+1} + \frac{1}{n} (-1)^{n+1}
+ \frac{1}{n} \sum_{q=0}^n B_q (-1)^{n-q} {n\choose q} 2^q
\\ = (-1)^{n+1} \frac{n+1}{n}
+ (n-1)! [z^n] \exp(-z) \frac{2z}{\exp(2z)-1}.$$
Now we have
$$\frac{2}{\exp(2z)-1} =
\frac{1}{\exp(z)-1} - \frac{1}{\exp(z)+1}$$
as well as
$$\frac{1}{\exp(z)} \frac{1}{\exp(z)-1}
= \frac{1}{\exp(z)-1} - \frac{1}{\exp(z)}$$
and
$$\frac{1}{\exp(z)} \frac{1}{\exp(z)+1}
= - \frac{1}{\exp(z)+1} + \frac{1}{\exp(z)}$$
which yields for the sum
$$(-1)^{n+1} \frac{n+1}{n}
+ (n-1)! [z^n]
\left[\frac{z}{\exp(z)-1} + \frac{z}{\exp(z)+1}
- \frac{2z}{\exp(z)} \right]$$
or
$$(-1)^{n+1} \frac{1-n}{n}
+ \frac{1}{n} B_n
+ (n-1)! [z^n] \frac{z}{\exp(z)+1}.$$
Observe that
$$\frac{z}{\exp(z)+1}
= \frac{z}{\exp(z)-1} - \frac{2z}{\exp(2z)-1}$$
This will produce at last
$$(-1)^{n+1} \frac{1-n}{n}
+ \frac{1}{n} B_n
+ \frac{1}{n} B_n - \frac{1}{n} 2^n B_n$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
(-1)^{n+1} \frac{1-n}{n} +
\frac{1}{n} (2-2^n) B_n.}$$
Again with the Bernoulli numbers at odd indices being zero we get for
$n\ge 3$, $n$ odd the closed form
$$\bbox[5px,border:2px solid #00A000]{
\frac{1-n}{n}.}$$
Here we have consulted OEIS A036968, the
Genocchi  numbers.
