Which functions besides $\ln{x}$ make $ \lim \limits_{N \to \infty} \sum_{n=1}^N f'(n) -f(N) $ converge? Using a very hand-wavy argument, I convinced myself that if, instead of $f(x)=\ln{x}$, we let $f(x)=\sqrt{x}$, we should still get something finite and small. Wasn't really sure where to start to prove it, so ran a program to see what happens for large N instead.
$$ L = \lim \limits_{N \to \infty} \sum_{n=1}^N \frac{1}{2\sqrt{n}}  -\sqrt{N} $$
It seems like $L$ approaches about $-0.73018$ but couldn't really tell if it wasn't just running away to negative infinity really really slowly. What tactics might we use to prove/disprove convergence here?
 A: Use the Euler-Maclaurin formula:
$$
\sum_{n=1}^Nf'(n)=\int_1^Nf'(t)\mathrm dt+\frac12f'(1)+\color{blue}{\frac12f'(N)+\int_1^N\rho(t)f''(t)\mathrm dt},
$$
where $\rho(t)=t-\lfloor t\rfloor-1/2$. Suppose $f'$ is continuously differentiable and decreases to zero. Then because $|\rho(t)|\le1/2$ and $f''$ is nonpositive, we see that $f''\le0$ and
$$
\int_1^N|\rho(t)|f''(t)\mathrm dt\le-\frac12\int_1^Nf''(t)\mathrm dt=\frac12\big(f'(1)-f'(N)\big)\le\frac12f'(1),
$$
so the blue term converges as $N\to\infty$. Therefore, we conclude the following result:
Theorem: Let $f:[1,+\infty)\mapsto[0,+\infty)$ be twice differentiable such that $f'$ is decreasing to zero and $f''$ is continuous. Then
$$
\lim_{N\to\infty}\left\{\sum_{n=1}^Nf'(n)-f(N)\right\}
$$
converges.
A: For the specific case, $L$, think of each term in a sequence
$$m_n = L_n-L_{n-1} = \frac{1}{2\sqrt{n}}-\sqrt{n}+\sqrt{n-1}=\frac{1-2n+2\sqrt{n(n-1)}}{2\sqrt{n}}$$
However, since $(n-\frac{1}{2})^2 = n(n-1) + \frac{1}{4} > n(n-1)$
$m_n < \frac{1 - 2n + 2(n-\frac{1}{2})}{2\sqrt{n}} = 0$ so the sequence is monotonically decreasing.
Then, since $f'(n)$ itself is strictly decreasing...
$$\int_{1}^Nf'(x)dx \leq \sum_{n=1}^Nf'(n)$$
This implies $L_N \geq -1$ and so the sequence is bounded from below.
Thanks TravorLZH for the more general answer to this and thanks to those who left a comment.
