# 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 $\frac15$.

The simplest solution (without the knowledge if the closed form formula) is to evaluate the Riemann sum:

$$\frac 1{n^5}\sum_{k=1}^n k^4 = \frac 1{n}\sum_{k=1}^n \left(\frac kn\right)^4 \to \int_0^1 x^4 dx$$

Hint: a Riemann sum for $\int_0^1 x^4\; dx$.

Well, what you're writing in the second line is basically T(n), the nth term. To try and find out the summation of the series try expanding the nth term. Having done that you'll find that it decomposes into a 4th power and lesser powers. Sum up the LHS and RHS. You'll automatically get what's going on.

You did want just a hint. Good Luck.