The limit of a series I need help with this exercise!
$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}.$
I saw somewhere online that $1^4+2^4+...+n^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.
But I dont understand why!
So following that then,
$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}=$
$=\lim\limits_{n \to \infty} \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}=$
$=\lim\limits_{n \to \infty} \dfrac{(n+1)(2n+1)(3n^2+3n-1)}{30n^4}=\dfrac15.$
And after that what!? Do I multiply and then use L'Hopital?
 A: This is what happens when you rush into learning calculus!
The quantity $S(n) = 1^4 + 2^4 + \cdots + n^4$ can be summed in a variety of ways. One such way is to guess that it grows like a fifth degree polynomial, then use Lagrange interpolation and induction. Another way might be to notice that
$$\sum_{i=0}^{n-1} (i+1)^5 - i^5 = n^5$$
is easy to evaluate. But alternatively,
$$(i+1)^5 - i^5 = 5i^4 + 10i^3 + 10i^2 + 5i + 1.$$
I leave it as an exercise to you on how to derive from here; now to your question on l'Hopital's.
You have $$\lim_{n\rightarrow\infty} \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}.$$
The numerator is a fifth degree polynomial. The denominator is a fifth degree polynomial. So the limit will just be the ratio of the leading coefficients, $6/30 = 1/5.$ Why? Well, imagine we have a simpler case,
$$\lim_{n\rightarrow\infty} \frac{n^2 + 3n + 1}{n^2}.$$
Then
$$\frac{n^2+3n+1}{n^2} = 1 + \frac{3}{n} + \frac{1}{n^2}.$$
As $n$ goes to infinity, anything with an $n$ in the denominator will go to 0, and you're just left with 1. If you expanded out that fifth degree numerator, you'd quickly find the same trick applies and all but the leading term dies.
A: Note $\sum_{k=1}^nk^4$ must be divisible by $n(n+1)$, so that the obvious extension to negative $n$ achieves $0-0^4=0$ at $n=-1$. The large-$n$ behaviour is asymptotic to $\int_0^nx^4dx=\tfrac15n^5$. We can determine coefficients in $\sum_{k=1}^nk^4=\tfrac15n(n+1)(n^3+An^2+Bn+C)$ from the sum at three values of $n$, say $n\in\{1,\,2,\,3\}$, thereby acquiring simultaneous equations. This gives $A=\tfrac32,\,B=\tfrac16,\,C=-\tfrac16$, after which the rational root theorem finds the factor $n+\tfrac12$.
