Find the limit of $\frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$ 
Let $\{x_n\}_{n\geq 1}$ be defined as $x_n = \frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$. Then $\lim\limits_{n\to\infty} x_n$ is?

I want to find limit of this problem by a very specific method . I am uploading the pic of that method and my attempt.
Please guide me as to how to take this method ahead. 

 A: Alternate solution:
$$x_n < \frac{n}{n^3+1} + \frac{2n}{n^3+1} + \cdots + \frac{n\cdot n}{n^3+1} = \frac{n(1+2+\cdots + n)}{n^3+1} =  \frac{n\cdot n(n+1)/2}{n^3+1}.$$
The limit of the last expression is $1/2.$ From below we have
$$\frac{n}{n^3+n} + \frac{2n}{n^3+n} + \cdots + \frac{n\cdot n}{n^3+n} = \frac{n\cdot n(n+1)/2}{n^3+n} < x_n.$$
The last fraction also has limit $1/2.$ By the squeeze theorem the desired limit is $1/2.$
A: Starting where you left: for $n\geq 1$,
$$
x_n = \frac{1}{n}\sum_{k=1}^n \frac{\frac{k}{n}}{1+\frac{k}{n^3}} \tag{1}
$$
While this furiously looks like a Riemann sum, it is not one due to the $\frac{k}{n^3}$ in the denominator. But then, we can use the squeeze theorem: as $1\leq k\leq n$,
$$
\frac{1}{1+\frac{1}{n^2}}\cdot \underbrace{\frac{1}{n}\sum_{k=1}^n \frac{k}{n}}_{}
= \frac{1}{n}\sum_{k=1}^n \frac{\frac{k}{n}}{1+\frac{1}{n^2}}
\leq x_n \leq  \frac{1}{n}\sum_{k=1}^n \frac{\frac{k}{n}}{1+\frac{1}{n^3}}
= \frac{1}{1+\frac{1}{n^3}}\cdot\underbrace{\frac{1}{n}\sum_{k=1}^n \frac{k}{n}}_{}
$$
and since $\lim_{n\to\infty}\frac{1}{1+\frac{1}{n^2}} = \lim_{n\to\infty} \frac{1}{1+\frac{1}{n^3}} = 1$, by the squeeze theorem the limit will be that of
$$
\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{k}{n} = \int_0^1 f(x)dx
$$
for some very simple function $f$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\lim_{n \to \infty}x_{n} & =
\lim_{n \to \infty}\sum_{k = 1}^{n}{kn \over n^{3} + k} =
\lim_{n \to \infty}\sum_{k = 1}^{n}{\pars{k + n^{3}}n - n^{4} \over k + n^{3}} =
\lim_{n \to \infty}\bracks{n^{2} -
n^{4}\sum_{k = 0}^{n - 1}{1 \over k + 1 + n^{3}}}
\\[5mm] & =
\lim_{n \to \infty}\bracks{n^{2} -
n^{4}\sum_{k = 0}^{\infty}
\pars{{1 \over k +1 + n^{3}} - {1 \over k + n + 1 + n^{3}}}}
\\[5mm] & =
\lim_{n \to \infty}\bracks{n^{2} - n^{4}\pars{H_{n + n^{3}} - H_{n^{3}}}}
\label{1}\tag{1}
\end{align}
where $\ds{H_{m}}$ is a Harmonic Number which has the
asymptotic expansion $\ds{\pars{\mbox{as}\ m \to \infty}}$:
$$
H_{m} \sim \ln\pars{m} + \gamma + {1 \over 2m} - {1 \over 12m^{2}}
$$
$\ds{\gamma}$ is the Euler-Mascheroni Constant.

Expression \eqref{1} becomes:
\begin{align}
\lim_{n \to \infty}x_{n} & =
\lim_{n \to \infty}\braces{n^{2} - n^{4}
\bracks{\ln\pars{n + n^{3} \over n^{3}} -
{1 \over 2}\,{n \over n^{3}\pars{n + n^{3}}}}} = \bbx{\ds{1 \over 2}}
\end{align}
