Calculating $\lim _{n\to \infty }\left(\frac{1\cdot n + 2\cdot(n-1) + 3\cdot (n-2)+ ... +1\cdot n}{n^2}\right)$?

Question

Hello everyone how can I calculate the limit of:

$\lim _{n\to \infty }\left(\frac{1\cdot n + 2\cdot(n-1) + 3\cdot (n-2)+ ... +1\cdot n}{n^2}\right)$?

My direction was to convert it to something looks like Riemann sum by doing this:

$\lim _{n\to \infty }\left(\frac{\sum_{k=0}^{n} (k+1)(n-k)}{n^2}\right)$

But I don't know how to continue.

Answer 1

The limit of that sequence is $\infty$, since\begin{align} \sum_{k=1}^nk(n-k+1)&=n\sum_{k=1}^nk-\sum_{k=1}^nk^2+\sum_{k=1}^nk\\&=n\frac{n(n+1)}2-\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2\\&=\frac{n(n+1)(n+2)}6,\end{align}and therefore$$\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^nk(n-k+1)=\lim_{n\to\infty}\frac{(n+1)(n+2)}{6n}=\infty.$$

Answer 2

You can finish the problem using the formula for sums of squares.

$$ \sum_{k=0}^n (k+1)(n-k) = n \sum_{k=0}^n k - \sum_{k=0}^n k^2 + \sum_{k=0}^n n -\sum_{k=0}^n k = \\ \frac{n^2 (n+1)}{2} - \frac{n(n+1)(2n+1)}{6} + n(n+1) - \frac{n(n+1)}{2} = \Theta(n^3) $$

So the limit will be infinity since the $n^3$ does not vanish.

Answer 3

You may notice that $\sum_{k=1}^{n}k(n+1-k)$ is a convolution, then apply stars&bars:

$$\begin{eqnarray*}\sum_{k=1}^{n}k(n+1-k) &=& [x^{n+1}]\left(\sum_{k\geq 1}kx^k\right)^2=[x^{n+1}]\frac{x^2}{(1-x)^4}\\&=&[x^{n-1}]\frac{1}{(1-x)^4}=\binom{n+2}{\color{red}{3}}.\end{eqnarray*} $$ This gives that $$ \lim_{n\to +\infty}\frac{1}{n^{\color{red}3}}\sum_{k=1}^{n}k(n+1-k) = \frac{1}{{\color{red}3}!}=\frac{1}{6} $$ but also without the first line it is pretty clear that $\sum_{k=1}^{n}k(n+1-k)$ is a cubic polynomial in the $n$ variable, so the given limit is $+\infty$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{n \to \infty }{1\cdot n + 2\cdot\pars{n - 1} + 3\cdot \pars{n - 2} + \cdots + 1\cdot n \over n^{2}}:\ {\large ?}}$

By repeatedly using [Stolz-Ces$\grave{\mrm{a}}$ro Theorem] (https://en.wikipedia.org/wiki/Stolz–Cesàro_theorem) : \begin{align} &\lim_{n \to \infty }{1\cdot n + 2\cdot\pars{n - 1} + 3\cdot \pars{n - 2} + \cdots + 1\cdot n \over n^{2}} \\[3mm] = & \lim_{n \to \infty }{\sum_{k = 1}^{n}k\pars{n - k + 1} \over n^{2}} = \lim_{n \to \infty }{n\sum_{k = 1}^{n}k - \sum_{k = 1}^{n}k\pars{k - 1} \over n^{2}} \\[3mm] = & \lim_{n \to \infty }{\bracks{\pars{n + 1}\sum_{k = 1}^{n + 1}k - n\sum_{k = 1}^{n}k} - \bracks{\sum_{k = 1}^{n + 1}k\pars{k - 1} - \sum_{k = 1}^{n}k\pars{k - 1}} \over \pars{n + 1}^{2} - n^{2}} \\[3mm] = & \lim_{n \to \infty }{\sum_{k = 1}^{n + 1}k \over 2n + 1} = \lim_{n \to \infty }{\sum_{k = 1}^{n + 2}k - \sum_{k = 1}^{n + 1}k \over \bracks{2\pars{n + 1} + 1} - \pars{2n + 1}} \\[3mm] = & \lim_{n \to \infty }{n + 2 \over 2} = \bbx{\color{red}{+\ \infty}} \end{align}