0

As the title states, I'm trying to prove that the sequence $a_n$ converges to $1/4$ where: $$a_n=\dfrac{1}{n^4}+\dfrac{2^3}{n^4}+\cdots+\dfrac{n^3}{n^4}$$

Is there any way to do this without relying on previous sum identities or integrals? I'm having trouble doing this with just the squeeze theorem.

Community
  • 1
Ayumu Kasugano
  • 363
  • The sum formula for consequtive cubes and/or realization of $a_n$ as a Riemann sum are both natural approaches. May I enquire as to why you want to use only the squeeze theorem specifically? Can you do, for example, the simpler $$\lim_{n\to\infty}(\frac1{n^2}+\frac2{n^2}+\cdots+\frac n{n^2})=\frac12$$ with squeeze alone? – Jyrki Lahtonen Apr 24 '17 at 05:20
  • Related : https://math.stackexchange.com/questions/478344/what-is-the-result-of-lim-n-to-infty-frac-sumn-i-1-iknk1 – Arnaud D. Oct 15 '19 at 17:54

4 Answers4

1

By the hockey-stick identity we have $$ \sum_{n=3}^{N}\binom{n}{3}=\binom{N+1}{4}\tag{1}$$ and it is trivial that $$ 6\binom{n+1}{3}\leq n^3 \leq 6\binom{n+2}{3} \tag{2}$$ hence $$ \frac{6}{N^4}\binom{N+2}{4}\leq \frac{1}{N^4}\sum_{n=1}^{N}n^3 \leq \frac{6}{N^4}\binom{N+3}{4} \tag{3}$$ and by squeezing the wanted limit is clearly $\frac{6}{24} = \color{red}{\large \frac{1}{4}}.$

Jack D'Aurizio
  • 353,855
0

You can use. Lt n-> infinite an+1/an .this should be less that 1. Once you show this. It means that the sequence is converging. Then the sum of cubes starting from 1 is (n(n+1)/2)^2 . With that we can show that it converges to 1/4

Lakshya Gupta
  • 99
  • 6
0

Prove $\sum k^3 =n^2(n+1)^2/4$ by induction (so it doesn't depend on previous results). Then it is easy.

marty cohen
  • 107,799
  • 2
    The same approach has been suggested by several other answerers (mostly deleted by now). The OP specifically disallowed the use of this formula as well as the realization of the general term as a Riemann sum of $\int_0^1x^3,dx$. A useful answer would then IMHO either give a different approach, or explain why none exists. – Jyrki Lahtonen Apr 24 '17 at 05:27
  • OP wrote "without relying on previous sum identities or integrals?" I am not replying on previous identities; I am proving the needed one right here – marty cohen Apr 24 '17 at 06:53
  • So your answer to the question "How can I do this without referring to result A?" is "Prove result A first as a part of your solution!" Is that your idea of a useful post? – Jyrki Lahtonen Apr 24 '17 at 07:20
  • It satisfies the condition. – marty cohen Apr 24 '17 at 07:27
0

Apply Stolz-Cesaro to

$$\tag 1 \frac{1^3 + 2^3 + \cdots + n^3}{n^4}.$$

We are led to

$$\frac{(n+1)^3}{(n+1)^4-n^4} = \frac{n^3 + 3n^2 + 3n + 1}{4n^3 + 6n^2 + 4n + 1} \to \frac{1}{4}.$$

By S-C, the limit of $(1)$ is $1/4.$

zhw.
  • 105,693