Is something wrong in my solution? We have $f:(-1,\infty)\rightarrow\mathbb{R},f(x)=\frac{x}{x+1}$ and need to show that $a_n=\sum_{k=1}^n f(k)-\int_0^n f(t)dt$ is bounded.
Here is all my steps:
$f'>0\Rightarrow f$ is strictly increasing. Therefore $f(c_k)>f(k)$
where $f(c_k)=\int_k^{k+1}f(x)dx,\forall c_k\in[k,k+1]$.
By the way: $$\sum_{k=1}^n f(k)<\int_1^{n+1}f(x)dx$$
$\Rightarrow a_n<\int_1^{n+1}(1-\frac{1}{x+1})dx-\int_0^n f(t)dt$

It is correct what I wrote of this?
How can I continue from here?

 A: $$
use \; :\int_{0}^{n}f(t)dt=\sum_{1}^{n}\int_{k-1}^{k}f(t)dt\; \; (1)\\then :f\;is \; increasing \;so:f(k-1)\leq f(t)\leq f(k)\\f(k-1)\leq \int_{k-1}^{k}f(t)dt\leq f(k)\\finally :0\leq a_{n}\leq \sum_{1}^{n}f(k)-f(k-1)=f(n)-f(0)\underset{\infty }{\rightarrow}1
$$
A: There is nothing wrong so far. Keep evaluating your right-hand side: $\int \limits _1 ^{n+1} f \Bbb \space d t - \int \limits _0 ^n f \Bbb \space d t = \int \limits _n ^{n+1} f \space \Bbb d t - \int \limits _0 ^1 f \space \Bbb d t = 1 + \ln {n \over n+1} - 1 + \ln 2 = \ln 2 + \ln {n \over n+1}$. Now note that $\frac 1 2 \leq \frac n {n+1} < 1$, so $\ln 2 + \ln {n \over n+1} \leq \ln 2 + \ln 1 = \ln 2$, which is an upper bound for your $a_n$.
A: $$a_n=\sum_{k=1}^{n}f(k)-\int_{0}^{n}f(t)\,dt = \sum_{k=1}^{n}\frac{1}{k+1}-\int_{0}^{n}\frac{dt}{t+1}\,dt=H_{n+1}-1-\log(n+1)$$
is negative because:
$$ \int_{k-1}^{k}\frac{dt}{t+1}=\int_{k}^{k+1}\frac{dt}{t}\in\left(\frac{1}{k+1},\frac{1}{k}\right) $$
since $g(t)=\frac{1}{t}$ is a positive decreasing function on $\mathbb{R}^+$. On the other hand:
$$ a_{n+1}-a_n = \frac{1}{n+2}-\log\left(1+\frac{1}{n+2}\right)\geq 0$$
gives that the sequence $\{a_n\}_{n\geq 1}$ is increasing, and the Hermite-Hadamard inequality gives:
$$\begin{eqnarray*} a_n &=& \sum_{k=1}^{n}\left(\frac{1}{k+1}-\int_{k}^{k+1}\frac{dt}{t}\right) \geq \frac{1}{2}\sum_{k=1}^{n}\left(\frac{1}{k+1}-\frac{1}{k}\right)\geq -\frac{1}{2}.\end{eqnarray*} $$
We have, indeed,
$$ \lim_{n\to +\infty}a_{n} = -1+\gamma = -0.422784335\ldots $$
where $\gamma$ is the Euler-Mascheroni constant.
