# Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ [closed]

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.

• Oct 19, 2017 at 14:11
• This gives a very short answer seee below: math.stackexchange.com/questions/718939/… Oct 19, 2017 at 15:03
• Oct 19, 2017 at 17:28
• @amWhy: Not the same question. We don't need the closed-form for the sum of 4th-powers to find the limit here. A slick solution is to use integrals to bound each of the numerator and denominator, and then the limit falls out immediately. =) Oct 20, 2017 at 7:11

Let $\ell\in \mathbb R$ be a finite number. If $$\lim _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=\ell ,\$$ then, the limit $${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\ell .\ }$$ This is the Stolz–Cesàro theorem. In your case $$a_n=\sum_{k=1}^n k^4, \quad b_n=\sum_{k=1}^{n+1} k^4$$ and $$a_{n+1}-a_n=(n+1)^4, \quad b_{n+1}-b_n=(n+2)^4,$$ so the limit is equal to 1.

• (+1) When reading the question, this is the idea that came to mind, only I used $\frac{a_n-a_{n-1}}{b_n-b_{n-1}}$.
– robjohn
Oct 19, 2017 at 15:35

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.

• @robjohn: I like to use Cesaro-Stolz as often as possible, but try to avoid L'Hospital's Rule as much as possible. I realized quite late that the fundamental difference between the two is that Cesaro-Stolz can be proved without using completeness but L'Hospital's Rule requires completeness so that Cesaro-Stolz is a simpler and effective tool. Oct 20, 2017 at 16:23

Hint: $$\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4} =\frac{1^4+2^4+\dots+n^4+(n+1)^4-(n+1)^4}{1^4+2^4+\dots+n^4+(n+1)^4}$$

• You must add that the denominator will be a polynomial with degree 5! Oct 19, 2017 at 13:53
• I mean for finding that the limit is $1$, you need degree of denominator is greater than 4 for second term to be zero. Well I may be wrong. Oct 19, 2017 at 14:03
• To sum up: "hint" of what?
– Did
Oct 20, 2017 at 15:24

If $a_n > 0, s_n = \sum_{k=1}^n a_n, r_n = s_{n+1}/s_n$, then, if $a_{n+1}/a_n \to 1$, $r_n \to 1$.

Proof: $r_n-1 =a_{n+1}/s_n$. Since $a_{n+1}/a_n \to 1$, then, for any fixed $m$, $a_{n+k}/a_n \to 1$ for $1 \le k \le m$.

Therefore, for any fixed $m$, $\dfrac{a_{n+1}}{s_n} \lt \dfrac{a_{n+1}}{\sum_{k=0}^{m-1} a_{n-k}} \to \dfrac1{m}$ so that $r_n-1 =\dfrac{a_{n+1}}{s_n} \to 0$.

Your case is $a_n = n^4$, and it is easy to show that $a_{n+1}/a_n \to 0$. This holds for $a_n = n^p$ for any fixed $p$.

Then $\dfrac{1^4+2^4+...+n^4+(n+1)^4}{1^4+2^4+...+n^4}=1+\dfrac{(n+1)^4}{O(n^5)}=1+O(\frac 1n)\to 1$ and so is the inverse of that.

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, $$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}= \lim_{n\to\infty}\frac{\frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.}{\frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.+(n+1)^4}=1$$