<s>[INST] 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.  
 [/INST] Very short Answer
From this What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression
we have that,
$$ \sum_{k=1}^{n} k^{4} = \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.$$
Therefore, By polynomial limit we have, 
$$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}= \lim_{n\to\infty}\frac{\frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.}{\frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.+(n+1)^4}=1$$
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