I have been trying to compute the following limit-
$$\lim_{n\to \infty} \dfrac {2021(1^{2020}+2^{2020}+3^{2020}….+n^{2020}) – n^{2021}}{2021(1^{2019}+2^{2019}+3^{2019}…..+n^{2019})} =L$$
By the Stolz-Cesàro theorem, letting the numerator and denominator be $a_n$ and $b_n$ respectively yields
$$L=\lim_{n\to \infty} n+1-\frac{1}{2021}\bigg((n+1)^2-n^2\left(\frac {n}{n+1}^{}\right)^{2019}\bigg)$$
Now, I don't see how binomial expansions(perhaps some other technique?) can help here. Any help would be appreciated.
Best Answer
By continuing your approach you can see the limit is $$ \begin{align} L &= \lim n+1 - \dfrac{1}{2021(n+1)^{2019}} \left( (n+1)^{2021} - (n+1 -1)^{2021} \right) \\ &= \lim n+1 - \dfrac{1}{2021(n+1)^{2019}}\left( \binom{2021}{1}(n+1)^{2020} - \binom{2021}{2} (n+1)^{2019} + \cdots \right) \\ &= \boxed{1010} \end{align}$$