Calculating $\lim _{n\to \infty }\left(\frac{1\cdot n + 2\cdot(n-1) + 3\cdot (n-2)+ ... +1\cdot n}{n^2}\right)$? 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.
 A: 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$$\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^nk(n-k+1)=\lim_{n\to\infty}\frac{(n+1)(n+2)}{6n}=\infty.$$
A: 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.
A: 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 $+\infty$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\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}
