# Limit of $(\sum_{k=0}^{n}k^4)/n^5$

So i was trying to find this limit:

$$\lim_{n\to\infty}\frac{ \sum_{k=0}^{n}k^4}{n^5}$$

which at first made me think it's zero but soon i realized that it's probably not. I tried expanding that but there's no $n^5$ in the explansion. Eventialy i tried something like

$$\sum_{k=0}^{n+1}k^4 - \sum_{k=0}^{n}k^4=(x+1)^4= An^4+Bn^3+Cn^2+Dn+E$$ But again this has no $n^5$ involved.. If someone could provide a hint..

• When looking at asymptotic performance, one can often consider a summation the same as an integral. Since $\int_0^n x^4\,dx = n^5/5$, we should expect that $\sum_0^n k^4 \approx n^5/5$. – apnorton Feb 9 '15 at 20:23
• @anorton $+ \mathcal O(n^4)$ ;) – AlexR Feb 9 '15 at 20:24

By Faulhaber's formula the numerator is $$\sum_{k=0}^n k^4 = \sum_{k=1}^n k^4 = \frac{6n^5 + 15n^4 + 10 n^3 - n}{30}$$ Wich makes the limit equal to $\frac15$.
$$\frac 1{n^5}\sum_{k=1}^n k^4 = \frac 1{n}\sum_{k=1}^n \left(\frac kn\right)^4 \to \int_0^1 x^4 dx$$
Hint: a Riemann sum for $\int_0^1 x^4\; dx$.