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Here I'm given this limit. $$\displaystyle \lim_{n \to \infty} \dfrac{\displaystyle \sum_{k=1}^n n^k}{\displaystyle \sum_{k=1}^n k^n}$$

$\displaystyle \sum_{k=1}^n n^k$ simplifies to $\dfrac{n(n^n-1)}{n-1}$ but I'm unable to tackle $\displaystyle \sum_{k=1}^n k^n$.

How do you evaluate this limit?

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  • $\begingroup$ No \displaystyle or \dfrac in the title, please. $\endgroup$ – Clayton Oct 5 '18 at 4:41
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    $\begingroup$ Just an idea, but you might be able to multiply/divide by $n^n$ for that sum and transform it into a Riemann sum. (I haven't thought it through, so it might be bad advice). $\endgroup$ – Clayton Oct 5 '18 at 4:44
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Note that $$ \sum_{k=0}^n k^n = \sum_{j=0}^n (n-j)^n = n^n \sum_{j=0}^n (1-j/n)^n$$ and using dominated convergence, $$ \sum_{j=0}^n (1-j/n)^n \to \sum_{j=0}^\infty e^{-j} = \frac{e}{e-1}$$ Thus $$ \frac{\sum_{k=0}^n n^k}{\sum_{k=0}^n k^n} \to \frac{e-1}{e}$$

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