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.
 A: Observe that 
$$
1^4+2^4+...+n^4 = \sum_{i=1}^n i^4 = \frac{1}{30}(6n^5+15n^4+10n^3-n)
$$ so if $k<5$ the limit is $+\infty$ (does not exist) but if $k>5$ the limit is $0$.
A: Do you know Riemann Sum?
$\begin{eqnarray}
&& \lim_{n\to\infty} \left( \frac{1^4+2^4+3^4+\ldots+n^4 }{n^k } \right) \\
&=& \lim_{n\to\infty}n^{4-k} \left( \left(\frac1n\right)^4 +\left(\frac2n\right)^4 + \left(\frac3n\right)^4 + \ldots + \left(\frac nn\right)^4 \right) \\
\end{eqnarray} $
Should it converge to a finite non-zero value, it should be in the form of $\displaystyle \lim_{n\to\infty} \frac1n \displaystyle\sum_{k=1}^n f\left( \frac kn \right) $
So $4-k = -1 \Rightarrow k = 5 $.
If you want to evaluate the limit: it becomes $ \displaystyle \int_0^1 x^4 \, dx = \frac15 $.

Alternatively, you can apply Stolz Cesaro Theorem: $a_n= 1^4 + 2^4 + \ldots+n^4 $, $ b_n = n^k $.
Show that $ \displaystyle \lim_{n\to\infty} = \frac{a_{n+1} - a_n}{b_{n+1} - b_n } = \lim_{n\to\infty} \frac{(n+1)^4}{(k+1)n^{k-1}} $, then we have $k-1 =4 $ again.
A: 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.
