# Evaluating the limit : $\lim_{n \to \infty} \frac{ \sum_{k=1}^n n^k}{ \sum_{k=1}^n k^n}$

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?

• No \displaystyle or \dfrac in the title, please. – Clayton Oct 5 '18 at 4:41
• 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). – Clayton Oct 5 '18 at 4:44

## 1 Answer

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 $$\frac{\sum_{k=0}^n n^k}{\sum_{k=0}^n k^n} \to \frac{e-1}{e}$$