We first prove that
$$\frac{x}{1+x} < \ln(1+x) < x, \quad x > 0. \qquad \textcircled{1}$$
Let
$$h(x) = x - \ln(1+x), \\ g(x) = \ln(1+x) - \frac{x}{1+x}.$$
Then, for $x > 0$,
$$h'(x) = 1 - \frac{1}{1+x} > 0, \\ g'(x) = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{x}{(1+x)^2} > 0.$$
Therefore,
$$h(x) > h(0) = 0, \quad g(x) > g(0) = 0.$$
This completes the proof of the inequalities ①.

Now let $x = \frac{1}{n}$ in ①. We have
$$\frac{1}{n+1} < \ln\left(1+\frac{1}{n}\right) < \frac{1}{n}. \qquad \textcircled{2}$$
Let
$$x_n = \sum_{k=1}^{n} \frac{k}{k^2 + 1} - \ln n.$$
Then
$$\begin{aligned} x_n - x_{n-1} &= \frac{n}{n^2+1} - \ln\left(1 + \frac{1}{n+1}\right) \\ &< \frac{n}{n^2+1} - \frac{1}{n} \\ &= -\frac{1}{n(n^2+1)} < 0. \end{aligned}$$
Therefore, $x_n < x_{n-1} < \cdots < x_1 = \frac{1}{2}$.

Furthermore,
$$\begin{aligned} \ln n &= (\ln n - \ln(n-1)) + (\ln(n-1) - \ln(n-2)) \\ &\quad + \cdots + (\ln 2 - \ln 1) + \ln 1 \\ &= \sum_{k=1}^{n-1} \ln\left(1 + \frac{1}{k}\right). \end{aligned}$$
Consequently,
$$\begin{aligned} x_n &= \sum_{k=1}^{n} \frac{k}{k^2+1} - \sum_{k=1}^{n-1} \ln\left(1 + \frac{1}{k}\right) \\ &= \sum_{k=1}^{n-1} \left( \frac{k}{k^2+1} - \ln\left(1 + \frac{1}{k}\right) \right) + \frac{n}{n^2+1} \\ &> \sum_{k=1}^{n-1} \left( \frac{k}{k^2+1} - \frac{1}{k} \right) \\ &= - \sum_{k=1}^{n-1} \frac{1}{(k^2+1)k} \\ &> - \sum_{k=1}^{n-1} \frac{1}{(k+1)k} \\ &= -1 + \frac{1}{n} > -1. \end{aligned}$$
This completes the proof.