Let $u_n=\dfrac{1+2^2+3^3+...+n^n}{n^n}$. Prove that : $$\lim u_n=1$$ I tried to prove that $u_n< something$ have limit is 1 ; but I can't
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
Divide and conquer: $$\frac{1+2^2+\dots + n^n}{n^n} < \frac{(n-2)(n-2)^{n-2}}{n^n} + \frac{(n-1)^{n-1}}{n^n} + 1 < \frac{2}{n} + 1$$ |
|||
|
|
|
You can do $$u_n<\frac{(n-2)^{n-1}+(n-1)^{n-1}+n^n}{n^n}<\frac 1n+\frac 1 n+1$$ The first term obtained by $r^r<(n-2)^{n-2}$ and adding the $n-2$ terms which result. The estimate is very crude. |
|||
|
|
$$\begin{align*} u_n-1&=\frac{1+2^2+3^3+\ldots+(n-1)^{n-1}}{n^n}\\ &\le\frac{1+2^2+\ldots+(n-2)^{n-2}}{n^n}+\frac{(n-1)^{n-1}}{n^n}\\ &\le\frac{(n-2)(n-2)^{n-2}}{n^n}+\frac1n\left(1-\frac1n\right)^{n-1}\\ &=\frac{(n-2)^{n-1}}{n^n}+\frac1n\left(1-\frac1n\right)^{n-1}\\ &=\frac1n\left(1-\frac2n\right)^{n-1}+\frac1n\left(1-\frac1n\right)^{n-1} \end{align*}$$ |
|||
|
|
