Find the limit of $\frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$

Question

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.

Answer 1

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 $f$.

Answer 2

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 $1/2.$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

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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}