Existence and value of $\lim_{n\to\infty} (\ln\frac{x}{n}+\sum_{k=1}^n \frac{1}{k+x})$ for $x>0$ Does the limit 
$$W(x)=\lim_{n\to\infty} \left(\ln\frac{x}{n}+\sum_{k=1}^n \frac{1}{k+x} \right)$$
exist for all $x>0$? If so, what is the limit
$$\lim_{x\to\infty}W(x)?$$

We have
$$\sum_{k=1}^n \dfrac1{k+x} = \int_{1^-}^{n^+} \dfrac{d \lfloor t \rfloor}{t+x} = \left. \dfrac{\lfloor t \rfloor}{t+x} \right\vert_{t=1^-}^{t=n^+} + \int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt = \dfrac{n}{n+x} + \int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt$$
Now
$$\int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt = \int_1^{n^+} \dfrac{t}{(t+x)^2} dt - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt$$
$$\int_1^{n^+} \dfrac{t}{(t+x)^2} dt = \int_1^{n} \dfrac{dt}{t+x} - x \int_1^n \dfrac{dt}{(t+x)^2} = \log(n+x) - \log(1+x) -x \left(\dfrac1{1+x} - \dfrac1{n+x}\right)$$
Hence, we get that
\begin{align}
\log(x/n) + \sum_{k=1}^n \dfrac1{k+x} & = \log\left(\dfrac{x}n \right) + \dfrac{n}{n+x} + \log \left(\dfrac{n+x}{1+x}\right) -\dfrac{x}{1+x} + \dfrac{x}{n+x} - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt\\
& = - \dfrac{x}{1+x} + 1 + \log\left(\dfrac{x}n \cdot \dfrac{n+x}{1+x}\right) - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt\\
& = \dfrac1{1+x} + \log\left(\dfrac{x}n \cdot \dfrac{n+x}{1+x}\right) - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt
\end{align}
Now letting $n \to \infty$, we get that
$$W(x) = \dfrac1{1+x} + \log \left(\dfrac{x}{1+x}\right) - \underbrace{\int_1^{\infty} \dfrac{\{t\}}{(t+x)^2} dt}_{\text{Converges since }\{t\} \in [0,1)}$$
$$0 \leq \overbrace{\int_1^{\infty} \dfrac{\{t\}}{(t+x)^2} dt}^{g(x)} \leq \int_1^{\infty} \dfrac1{(t+x)^2} dt = \dfrac1{1+x}$$ There might be some name for $g(x)$ (Probably some of the number theorists on this website might be able to identity this). For instance, $g(0) = 1-\gamma$, where $\gamma \approx 0.57721$ is the Euler Mascheroni constant.
Now $$\lim_{x \to \infty} W(x) = 0 + \log(1) + 0 = 0$$

Another method is as follows. From here, we have
\begin{align}
\sum_{k=1}^n \left(\dfrac1k - \dfrac1{x+k} \right) & = \sum_{k=1}^n \int_0^1 (y^{k-1} - y^{x+k-1})dy\\
& = \int_0^1 (1-y^x) \sum_{k=1}^n y^{k-1} dy\\
& = \int_0^1 (1-y^x) \dfrac{1-y^n}{1-y} dy
\end{align}
Hence, we have
\begin{align}
\log(x/n) + \sum_{k=1}^n \dfrac1{k+x} & = \log(x/n) + \sum_{k=1}^n \left(\dfrac1{k+x} - \dfrac1k \right) + \sum_{k=1}^n \dfrac1k\\
& = \log(x) + \sum_{k=1}^n \dfrac1k - \log(n) + \sum_{k=1}^n \left(\dfrac1{k+x} - \dfrac1k \right)\\
& = \log(x) + \sum_{k=1}^n \dfrac1k - \log(n) - \int_0^1 (1-y^x) \dfrac{1-y^n}{1-y} dy
\end{align}
Now letting $n \to \infty$, we get that
$$W(x) = \log(x) + \gamma - \int_0^1 \dfrac{1-y^x}{1-y} dy$$
Now as $x \to \infty$, we have $$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$ Hence, we get that
$$\lim_{x \to \infty} W(x) = 0$$

Let us prove why, as $x \to \infty$, we have $$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$
The proof is the same as before. We have
$$\sum_{k=1}^n \dfrac1k = \int_0^1 \sum_{k=1}^n y^{k-1} dy = \int_0^1 \dfrac{1-y^n}{1-y} dy$$
But we know that $\displaystyle \sum_{k=1}^n \dfrac1k = \log(n) + \gamma + \mathcal{O}(1/n)$. Hence, we get that
$$\int_0^1 \dfrac{1-y^n}{1-y} dy = \log(n) + \gamma + \mathcal{O}(1/n)$$
 Replacing $n$ by $x$ and because the integral is a smooth function of $x$, we can conclude that
$$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$
