Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ I am just trying to calculate 
$$\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}.$$
To do this I apply formula for sum of fourth powers of $n$ number. My result: $$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}=1$$. I'm intrested in finding other method to solve the following problem.  

As remarked by SBA, by denoting $1^4+2^4+\ldots+n^4$ as $p(n)$ we have
$$ \lim_{n\to +\infty}\frac{p(n)}{p(n+1)} \stackrel{\text{Stoltz-Cesàro}}{=}\lim_{n\to +\infty}\frac{n^4}{(n+1)^4} = 1 $$
but the same conclusion simply follows from the observation that $p(n)$ is a polynomial.
