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I stuck to do this,

$$\lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}=1.$$

The only thing I have observed is $$ 1\le \lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}$$ I am unable to get its upper estimate so that I can apply Sandwich's lemma.

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  • $\begingroup$ Cesaro-Stolz should do the trick $\endgroup$ Sep 11 '17 at 20:43
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$$1\le \frac{1+2^2+\ldots+n^n}{n^n}\le \frac{n+n^2+\ldots+n^n}{n^n} = \frac{n\frac{n^n-1}{n-1}}{n^n} = \frac{n^{n+1}-n}{n^{n+1}-n^n}\xrightarrow{n\to\infty} 1$$

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  • $\begingroup$ Why $ \frac{n^{n+1}-n}{n^{n+1}-n^n}\xrightarrow{n\to\infty} 1$? $\endgroup$ Sep 11 '17 at 21:01
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    $\begingroup$ @SachchidanandPrasad Try dividing each term in the entire fraction by the highest degree term - in this case $n^{n+1}$. It should follow without much pain from there. $\endgroup$
    – Ducky
    Sep 11 '17 at 21:03
  • $\begingroup$ Okay okay, sorry it was easy. Thanks $\endgroup$ Sep 11 '17 at 21:04
  • $\begingroup$ It might be clearer to end with $${n{n^n-1\over n-1}\over n^n}=\left(n\over n-1\right)\left(1-{1\over n^n}\right)\to1$$ $\endgroup$ Sep 12 '17 at 2:29
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Consider any sequence $a_n$ with the property that $\lim\limits_{n\to\infty}a_n=\infty$ and $\lim\limits_{n\to\infty}\frac{a_{n-1}}{a_n}=0$. It follows by Stolz–Cesàro that

$$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{a_n}=\lim_{n\to\infty}\frac{a_n}{a_n-a_{n-1}}=\lim_{n\to\infty}\frac1{1-\frac{a_{n-1}}{a_n}}=1$$

Here, we have $a_n=n^n$.

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$$1+2^2+3^3+\cdots(n-2)^{n-2}+(n-1)^{n-1}+n^n<\\n^{n-2}+n^{n-2}+n^{n-2}+\cdots n^{n-2}+n^{n-1}+n^n.$$

When divided by $n^n$, these terms are bounded by $\dfrac n{n^2}+\dfrac1n+1$.

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$$\lim_{n \rightarrow \infty} \frac{1^1 + 2^2 + \dots + (n-1)^{n-1} + n^n}{n^n} = \lim_{n \rightarrow \infty} \left(\frac{1^1}{n^n} + \frac{2^2}{n^n} + \dots + \frac{(n-1)^{n-1}}{n^n} + 1 \right) $$

Let's examine that second to last term:

$$ \begin{align} \lim_{n \rightarrow \infty} \frac{(n-1)^{n-1}}{n^n} &= \lim_{n \rightarrow \infty} \frac{1}{n}\frac{(n-1)^{n-1}}{n^{n-1}} \\ &= \lim_{n \rightarrow \infty} \frac{1}{n} \left( \frac{n-1}{n} \right)^{n-1}< \lim_{n \rightarrow \infty} \frac{1}{n} \cdot 1\\ &= 0 \end{align}$$ The second largest term is zero, none of the terms are negative, the sum is 1: $$\lim_{n \rightarrow \infty} \frac{\sum\limits_{m=1}^{n} m^m}{n^n} = 1$$

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