Limits. Finding positive value

Question

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.

Answer 1

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 $0$.

Answer 2

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 $.

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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.

Answer 3

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.