The quantity I'm interested in is expressed as follows:

$$ I = \mathbb{E}_{k\sim \text{Binom}(n,p)} \left[\ln \frac{\text{Beta}(p;a+k,b+n-k)}{\text{Beta}(p;a,b)}\right] $$

The term inside the expectation can be interpreted as the log ratio between the posterior and the prior of the Beta distribution evaluated at $p$ after observing $n$ Bernoulli($p$) coin flips with $k$ heads. As $n$ grows, I suspect that this expectation should grow no slower than $\ln(n)$ regardless of the choice for $a,b>0$ and $p\in[0,1]$. In short, I want to show that $I=\Omega(\ln(n))$.

**My attempt**

Recall the p.d.f of the Beta distribution at $p$ is:

$$ \begin{align} \text{Beta}(p;a,b)&=p^{a-1}(1-p)^{b-1} B(a,b)^{-1}\\ &=p^{a-1}(1-p)^{b-1} \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} \end{align} $$

We rewrite the quantity of interest $I$ as: $$ \begin{align} I &= \mathbb{E}_{k\sim \text{Binom}(n,p)} \left[\ln \frac{p^k(1-p)^{n-k}\Gamma(a+b+n)\Gamma(a)\Gamma(b)}{\Gamma(a+k)\Gamma(b+n-k)\Gamma(a+b)}\right]\\ &= np\ln(p)+n(1-p)\ln(1-p) + \mathbb{E}_{k\sim \text{Binom}(n,p)} \left[\ln \frac{\Gamma(a+b+n)\Gamma(a)\Gamma(b)}{\Gamma(a+k)\Gamma(b+n-k)\Gamma(a+b)}\right] \end{align} $$

At this point, I'm not sure how to move forward with the expectation of the Gamma terms.