<s>[INST] 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?
 [/INST] 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 $n+\tfrac12$.
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