<s>[INST] Limits. Finding positive value Find a real number k such that the limit $$\lim_{n\to\infty}\ \left(\frac{1^4 + 2^4 + 3^4 +....+ n^4}{n^k}\right)$$ has as positive value. 
If I am not mistaken every even $k$ can be the answer. But the answer is 5.
 [/INST] First note that
$$\sum\limits_{i=1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
$$=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}$$
So now we have
$$\lim\limits_{n\to\infty} \left(\frac{1^4 + 2^4 + 3^4 +\cdots + n^4}{n^k}\right)$$
$$=\lim\limits_{n\to\infty} \left(\frac{\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}}{n^k}\right)$$
$$=\lim\limits_{n\to\infty} \left(\frac{n^5}{5n^k}+\frac{n^4}{2n^k}+\frac{n^3}{3n^k}-\frac{n}{30n^k}\right)$$
If $k\lt 5$, then
$$\lim\limits_{n\to\infty} \left(\frac{n^5}{5n^k}+\frac{n^4}{2n^k}+\frac{n^3}{3n^k}-\frac{n}{30n^k}\right)=\infty$$
If $k\gt 5$, then
$$\lim\limits_{n\to\infty} \left(\frac{n^5}{5n^k}+\frac{n^4}{2n^k}+\frac{n^3}{3n^k}-\frac{n}{30n^k}\right)=0$$
If $k=5$, then
$$\lim\limits_{n\to\infty} \left(\frac{n^5}{5n^5}+\frac{n^4}{2n^5}+\frac{n^3}{3n^5}-\frac{n}{30n^5}\right)$$
$$=\lim\limits_{n\to\infty} \left(\frac{1}{5}+\frac{1}{2n}+\frac{1}{3n^2}-\frac{1}{30n^4}\right)=\frac15$$
Therefore the answer is $k=5$ because zero is not positive nor negative and infinity is not quantifiable.
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