Question

The value of $\underset{n\to \infty }{lim}\frac{1^1+2^2+3^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}$ is

• A. $\frac{e-1}{e}$
• B. $\frac{e}{e-1}$
• C. $\frac{1}{e-1}$
• D. $\frac{e-1}{e+1}$

Answer 1

The correct option is A $\frac{e-1}{e}$

$L=\underset{n\to \infty }{lim}\frac{1^1+2^2+3^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}$

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

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

$=\underset{n\to \infty }{lim}\frac{1}{1+{(1-\frac{1}{n})}^n+{(1-\frac{2}{n})}^n+{(1-\frac{3}{n})}^n+\cdots }$

$=\frac{1}{1+e^{-1}+e^{-2}+e^{-3}+\cdots }\left[\text{If}\underset{x\to a}{lim}f\right(x)=1\&\underset{x\to a}{lim}g(x)=\infty ,\text{then}\underset{x\to a}{lim}\left[f\right(x){]}^{g\left(x\right)}=e{}^{\underset{x\to a}{lim}\left[f\right(x)-1]g\left(x\right)}]$

$=\frac{1}{\frac{1}{1-e^{-1}}}$
$=\frac{e-1}{e}$