Finding the limit to infinity of a summation The following problem was featured as a challenge in a previous exam paper and has left me stumped.
Compute the following limit:
$$
\lim_{n \to \infty} \frac{1}{n^{2013}} \sum_{k=1}^n k^{2012}
$$
Any help or hints to get me on the right track would be greatly appreciated!
 A: Two approaches.
(I)
$$ 2012!\binom{k}{2012}\leq k^{2012} \leq 2012!\binom{k+2012}{2012}, $$
hence:
$$ 2012!\binom{n+1}{2013}\leq \sum_{k=1}^{n}k^{2012} \leq 2012!\binom{n+2013}{2013} $$
and if we divide both sides by $n^{2013}$ and take the limit as $n\to +\infty$ we get that the limit is $\color{red}{\large\frac{1}{2013}}$ by squeezing.
(II)
By Riemann sums,
$$ \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^{2012} = \int_{0}^{1}x^{2012}\,dx = \color{red}{\frac{1}{2013}}.$$
A: Hint:
This is kind of complicated, so the first thing you want to do is find a pattern somehow. Let's say the general form of the limit is the following:
$$\lim_{n \to \infty} \frac 1 {n^{m+1}} \sum_{k=0}^n k^m$$
The limit you want to find is $m=2012$, but start with $m=2,3,4,5,...$ first to find a pattern. If you need help finding a formula for the $\sum_{k=0}^n k^m$ part, here are the formulas for this sum from $m=1$ to $m=10$.
A: The Stolz–Cesaro theorem is one of most standard methods for such examples.
See
https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem
or with proof:
https://ru.wikipedia.org/wiki/%D0%A2%D0%B5%D0%BE%D1%80%D0%B5%D0%BC%D0%B0_%D0%A8%D1%82%D0%BE%D0%BB%D1%8C%D1%86%D0%B0
For $b_n=n^{2013}$ and $a_n=\sum_{k=1}^{n}k^{2012}$ we have the case when it works.
$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{(n+1)^{2012}}{(n+1)^{2013}-n^{2013}}=\frac{(n+1)^{2012}}{(n+1)^{2012}+(n+1)^{2011}n^{1}+\dots}$$
divide by $(n+1)^{2012}$ to find the limit: $\frac{1}{2013}$.
Of course you can generalize for other exponents instead of 2013, we have year 2016 currently :-]
