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.

  • 2
    $\begingroup$ Why did you delete math.stackexchange.com/q/4042052/42969 and post the identical question again? $\endgroup$
    – Martin R
    Feb 28 at 6:30
  • 1
    $\begingroup$ You should be able to note that the problem is much much simpler to handle if replace $2020$ by $a$, $2019$ by $b$, $2021$ by $c$ and use $b=a-1,c=a+1$ suitably wherever needed. $\endgroup$ Feb 28 at 12:28

By continuing your approach you can see the limit is $$ \begin{align} L &= \lim n+1 - \dfrac{1}{2021(n+1)^{2019}} \left( (n+1)^{2021} - (n+1 -1)^{2021} \right) \\ &= \lim n+1 - \dfrac{1}{2021(n+1)^{2019}}\left( \binom{2021}{1}(n+1)^{2020} - \binom{2021}{2} (n+1)^{2019} + \cdots \right) \\ &= \boxed{1010} \end{align}$$


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