aggierabz2006zw

2022-07-13

Finding $\underset{n\to \mathrm{\infty }}{lim}\frac{1+{2}^{2}+{3}^{3}+{4}^{4}+\cdots +{n}^{n}}{{n}^{n}}$
Finding
$\underset{n\to \mathrm{\infty }}{lim}\frac{1+{2}^{2}+{3}^{3}+{4}^{4}+\cdots +{n}^{n}}{{n}^{n}}$
Attempt:
$\underset{n\to \mathrm{\infty }}{lim}\left[\frac{1}{{n}^{n}}+\frac{{2}^{2}}{{n}^{n}}+\frac{{3}^{3}}{{n}^{n}}+\cdots \cdots +\frac{{n}^{n}}{{n}^{n}}\right]=1$
because all terms are approaching to zero except last terms
but answer is not $1$ , could some help me to solve it , thanks

Giovanna Erickson

Expert

Bounding by a geometric series,
$\begin{array}{rl}& \frac{{n}^{n}}{{n}^{n}}+\frac{\left(n-1{\right)}^{n-1}}{{n}^{n}}+\frac{\left(n-2{\right)}^{n-2}}{{n}^{n}}+\cdots +\frac{{1}^{1}}{{n}^{n}}\\ & \le 1+\frac{1}{n}+\frac{1}{{n}^{2}}+\frac{1}{{n}^{3}}+\cdots \\ & =\frac{n}{n-1}\end{array}$
Since the sum is obviously always $\ge 1$, and $\le \frac{n}{n-1}$, the Squeeze Theorem says that the limit is $1$

Jamison Rios

Expert

By Stolz
$\underset{n\to \mathrm{\infty }}{lim}\frac{1+{2}^{2}+{3}^{3}+{4}^{4}+\cdots +{n}^{n}}{{n}^{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{{n}^{n}}{{n}^{n}-\left(n-1{\right)}^{n-1}}=1$

Do you have a similar question?