Let
$$
x_n=\sum_{k=0}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx
$$
We will use the following result:
Lemma If $g:[0,1]\to\mathbb{R}$ is a continuously differentiable function. Then
$$
\frac{g(0)+g(1)}{2}-\int_0^1g(x)dx= \int_{0}^1\left(x-\frac{1}{2}\right)g'(x)dx.
$$
Indeed, this is just integration by parts:
$$\eqalign{
\int_{0}^1\left(x-\frac{1}{2}\right)g'(x)dx
&=\left.\left(x-\frac{1}{2}\right)g(x)\right]_{x=0}^{x=1}
-\int_0^1g(x)dx\cr
&=\frac{g(1)+g(0)}{2}-\int_0^1g(x)dx
}$$
Now applying this to the functions
$x\mapsto f\left(\frac{k+x}{n}\right)$ for $k=0,1,\ldots,n-1$ and adding the resulting inequalities we obtain
$$
x_n-\frac{f(0) +f(1)}{2} = \int_0^1\left(x-\frac{1}{2}\right)H_n(x)dx\tag{1}
$$
where,
$$
H_n(x)=\frac{1}{n}\sum_{k=0}^{n-1}f'\left(\frac{k+x}{n}\right)
$$
Clearly for every $x$, $H_n(x)$ is a Riemann sum of the function continuous $f'$, hence
$$
\forall\,x\in[0,1],\quad\lim_{n\to\infty}H_n(x)=\int_0^1f'(t)dt
$$
Moreover, $| H_n(x)|\leq\sup_{[0,1]}|f'|$. So,
taking the limit in $(1)$ and
applying the Dominated Convergence Theorem, we obtain
$$
\lim_{n\to\infty}\left(x_n-\frac{f(0) +f(1)}{2}\right)=
\left(\int_0^1f'(t)dt\right)\int_0^1\left(x-\frac{1}{2}\right)dx=0.
$$
This proves that
$$
\lim_{n\to\infty}x_n=\frac{f(0) +f(1)}{2}
$$
And consequently
$$
\lim_{n\to\infty}\left(\sum_{k=\color{red}{1}}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx\right)=\frac{f(1)-f(0)}{2}
$$
The logarithm of the initial expression is
$$
\frac 1n \sum_{k=1}^n \ln\left(\frac{k^2+kn}{n^2}\right) = \frac 1n \sum_{k=1}^n \ln\left(\left(\frac kn \right)^2 + \frac kn \right)
$$
which is a Riemann sum for
$$
\int_0^1 \ln(x^2 + x) \, dx = \int_0^2 \ln x \, dx = 2 \ln 2 - 2 \, .
$$
Best Answer
Consider $$\frac{1}{n}\sum_1^n \left(\frac{k}{n}\right)^{15}.\tag{1}$$ This is a right Riemann sum for $$\int_0^1 x^{15}\,dx.$$ The limit as $n\to\infty$ of (1) exists. From that, you should be able to find the answers to both questions.
Remark: We used a Riemann sum, since that seemed to be the approach requested. Another way of viewing things is that $\sum_1^n k^{15}$ is a polynomial in $n$ of degree $16$. Thus the first sequence obviously diverges to $\infty$, and the second converges to $0$.