The discussion revolves around the limit of the sum of powers divided by the highest power as n approaches infinity. Participants explore whether this limit can be evaluated using Riemann sums, focusing on the mathematical reasoning behind the expression.
Participants have not reached a consensus on the method for proving the limit, and multiple approaches, including the use of Riemann sums, are being considered.
The discussion does not clarify the assumptions or definitions necessary for evaluating the limit, nor does it resolve the mathematical steps involved in the proposed approaches.
Thankyou, June29, for your participation. Well done!(Nod)June29 said:Call the limit $\ell$. Obviously $1 \leqslant \ell $. For any $n,k \in \mathbb{N}\setminus\left\{0\right\}$ s.t. $k \leqslant n$ we have $k^k \leqslant n^k$, so we have:
$\displaystyle 1 \leqslant \ell \leqslant \lim_{{n}\to{\infty}}\frac{n^1+n^2+n^3+...+(n-1)^{n-1}+n^n}{n^n} = \lim_{n \to \infty} \frac{1}{n^n} \cdot \frac{n^{n+1}-n}{n-1} =\lim_{n \to \infty} \frac{1- {1}/{n^{n}}}{1-1/n} = 1. $
Thus $\ell = 1$. (First equality is geometric series). I'd love to know if it can be done via Riemann sums.
lfdahl said:Thankyou, June29, for your participation. Well done!(Nod)
Maybe, we should continue this thread, and ask in the forum, if the limit can be found by means of Riemanns sums?