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?
The notation "lim n→∞" stands for the limit as n approaches infinity. In other words, we are looking at the behavior of the expression as n gets larger and larger.
To prove this equation, we can use the definition of a limit. We need to show that for any small positive number ε, there exists a corresponding positive integer N such that when n is greater than N, the absolute difference between the expression and 1 is less than ε.
The increasing exponent represents the fact that each term in the numerator is getting larger and larger as n increases. This means that the sum of all these terms is approaching infinity, while the denominator (n^n) is also approaching infinity. Therefore, the limit of the expression is equal to 1.
Yes, we can use mathematical induction to prove this equation. We can first prove the base case (n=1), which is trivial since 1^1=1 and the expression becomes 1/1=1. Then, we can assume the equation holds for some positive integer k and prove that it also holds for k+1. This will show that the equation is true for all positive integers, including infinity.
Yes, there are multiple ways to prove this equation. Aside from using mathematical induction, we can also use the squeeze theorem, which states that if two functions have the same limit as x approaches infinity, and a third function is always between them, then the third function also has the same limit. In this case, we can use the functions 1 and n^n as the bounds, and show that the expression lies between them.