Interesting Limit (13)

$\lim_{n\to \infty} \dfrac{1^{2018} +2^{2018} +\cdots n^{2018} }{n^{2018}}-\dfrac{n}{2019} =\lim_{n\to \infty}\dfrac{2019\left(1^{2018} +2^{2018} +\cdots +n^{2018}\right)-n^{2019}}{2019n^{2018}}=\dfrac{\infty}{\infty}$ Let $a_n =2019\left(1^{2018} +2^{2018} +\cdots +n^{2018}\right)-n^{2019}$ and $b_n = 2019n^{2018}$ . To evalute the limit we apply Stolz-Cesàro Theorem 1 . Then $\lim_{n\to \infty} \dfrac{a_n}{b_n} =\lim_{n\to \infty} \dfrac{a_{n+1} -a_n}{b_{n+1} -b_n}=\lim_{n\to \infty} \dfrac{2019(n+1)^{2018}-(n+1)^{2019} + n^{2019}}{2019(n+1)^{2018}- 2019n^{2018}} \\ \lim_{n\to \infty} \dfrac{a_n}{b_n} = \boxed{\dfrac{1}{2}}$

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Note: $\begin{aligned} L & = \lim_{n\to \infty} \dfrac{2019\displaystyle \sum_{r=0}^{2018}\binom{2018}{r}n^{2018-r} +n^{2019} -\displaystyle\sum_{k=0}^{2019}\binom{2019}{k}n^{2019-k}}{2019\displaystyle\sum_{m=0}^{2018}\binom{2018}{m}n^{2018-m}-2019n^{2018}} \\ L& = \lim_{n\to \infty}\dfrac{2019\left(n^{2018} +2018n^{2017} +\dfrac{2018!}{2!.2016!}n^{2016}+\cdots +1\right )+ n^{2019} -\left(n^{2019}+2019n^{2018} + \dfrac{2019!}{2!.2017!} n^{2017}+\cdots +1\right) }{2019\left(n^{2018} +2018n^{2017} +\dfrac{2018!}{2!.2016!}n^{2016}+\cdots +1 \right)-2019n^{2018}} \\ L & = \lim_{n\to \infty}\dfrac{2018.2019n^{2017} + \cdots +2019 -\dfrac{2019!}{2!.2017!}n^{2017} - \cdots -1}{2019.2018 n^{2017} + \dfrac{2018!}{2!.2017!} n^{2016}+\cdots +1} =\dfrac{2019.2018 - \dfrac{2019!}{2!.2017!}}{2019.2018}=\dfrac{2019.2018\left(1-\dfrac{1}{2}\right)}{2019.2018}=\boxed{ \dfrac{1}{2} }\end{aligned}$

The summation formulas for the first several powers of $k$ are

$\begin{aligned} \sum_{k=1}^n k \ \ &= \frac{n(n+1)}2 = \frac{n^2}{2} + \frac{n}{2} \\ \sum_{k=1}^n k^2 &= \frac{n(n+1)(2n+1)}6 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \\ \sum_{k=1}^n k^3 &= \frac{n^2(n+1)^2}{4} = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4} \\ \sum_{k=1}^n k^4 &= \frac{n(n+1)(2n+1)(3n^2+3n+1)}{30} = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30} \\ \sum_{k=1}^n k^5 &= \frac{n^2(n+1)^2(2n^2+2n-1)}{12} = \frac{n^6}{6} + \frac{n^5}{2} + \frac{5n^4}{12} - \frac{n^2}{12} \end{aligned}$

From the above we notice that the first two terms in the formula for $\displaystyle \sum_{k=1}^n k^a$ appear to be $\dfrac{k^{a+1}}{a+1} + \dfrac{k^a}{2}$ , and the remaining terms are $O \left( k^{a-1} \right)$ .

This is indeed the case, which can be proven with Faulhaber's formula, written about in the wiki Sum of n, n², or n³ .

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Armed with this result, we tackle the given limit:

$\begin{aligned} &\quad \lim_{n \to \infty} \frac{1^{2018} + 2^{2018} + 3^{2018} + .... + n^{2018}}{n^{2018}} - \frac{n}{2019} \\ \\ &= \lim_{n \to \infty} \frac{ \frac{n^{2019}}{2019} + \frac{n^{2018}}{2} + O \left( n^{2017} \right) }{n^{2018}} - \frac{n}{2019} \\ \\ &= \lim_{n \to \infty} \left( \frac{n}{2019} + \frac{1}{2} + O \left( \frac{1}{n} \right) \right) - \frac{n}{2019} = \boxed{\dfrac{1}{2}} \end{aligned}$