# limit: $u_n=\frac{1+2^2+3^3+…+n^n}{n^n}$.

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

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perhaps you can use links in this – Maesumi Jan 7 '13 at 16:06

Divide and conquer: $$\frac{1+2^2+\dots + n^n}{n^n} < \frac{(n-2)(n-2)^{n-2}}{n^n} + \frac{(n-1)^{n-1}}{n^n} + 1 < \frac{2}{n} + 1$$

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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.

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Another possibility is to replace the first terms of the numerator by $(n-1)+(n-1)^2 + \dots (n-1)^{n-1}$ which would be worth trying if the crude estimate did not work. The intuition for the crude version is that the function $n^n$ increases very rapidly - faster e.g. than $n!$ and this multiplicative growth is enough to beat adding together the earlier terms. – Mark Bennet Jan 7 '13 at 16:19

\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*}

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