Series inequality induction proof

In summary: Can you clarify the steps you took to evaluate both sides? Also, could you provide the instructions for the proof by induction?
  • #1
Aristarchus_
95
7
Homework Statement
How do I show that ##1+2^{2} + 3^{3} +...+n^{2} > \frac {1}{3} \cdot n^{3}## ?
Relevant Equations
k
My first attempt was ##... + n^{2} + (n+1)^{2} > \frac {1}{3} n^{3} + (n+1)^{2}##
then we must show that ##\frac {1}{3} n^{3} + (n+1)^{2} > \frac {1}{3} (n+1)^{3}##

We evaluate both sides and see that the LHS is indeed bigger than RHS. However, this solution is inconsistent so I am asking for some guidance as to a better method...
 
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  • #2
What you have done does not look right as an attempt at an inductive proof.

Please make a genuine attempt at a proof by induction.
 
  • #3
Is there a typo in your statement of the problem? In ##1+2^2+3^3+...+n^2##, I fail to see a pattern in the exponents. Is ##3^3## correct? Is ##n^2## correct?
 
  • #4
Aristarchus_ said:
Homework Statement:: How do I show that ##1+2^{2} + 3^{3} +...+n^{2} > \frac {1}{3} \cdot n^{3}## ?
Relevant Equations:: k

My first attempt was ##... + n^{2} + (n+1)^{2} > \frac {1}{3} n^{3} + (n+1)^{2}##
then we must show that ##\frac {1}{3} n^{3} + (n+1)^{2} > \frac {1}{3} (n+1)^{3}##

We evaluate both sides and see that the LHS is indeed bigger than RHS. However, this solution is inconsistent so I am asking for some guidance as to a better method...

Assuming you mean [tex]
1+2^{2} + 3^{2} + \dots +n^{2} > \frac {1}{3} n^{3}
[/tex] I would use [tex]
1 + 2^2 + 3^2 + \dots + n^2 = \tfrac16n(n+1)(2n+1).[/tex] But if I was required to prove that result rather than just state it, then I would prefer your approach since it requires slightly less work.
 
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  • #5
Aristarchus_ said:
We evaluate both sides and see that the LHS is indeed bigger than RHS. However, this solution is inconsistent so I am asking for some guidance as to a better method...
What do you mean by the solution is inconsistent?
 

1. What is series inequality induction proof?

Series inequality induction proof is a mathematical technique used to prove that a series of numbers or terms follows a certain pattern or inequality. It involves using mathematical induction, which is a method of proving that a statement is true for all natural numbers, to show that the desired inequality holds for each term in the series.

2. How is series inequality induction proof different from other types of mathematical proofs?

Series inequality induction proof is different from other types of mathematical proofs because it specifically focuses on proving inequalities within a series. Other types of proofs may focus on proving equalities or other types of mathematical statements.

3. What is the process for conducting a series inequality induction proof?

The process for conducting a series inequality induction proof involves three main steps. First, you must establish a base case, which is typically the first term in the series. Then, you assume that the inequality holds for a specific term in the series and use mathematical induction to show that it also holds for the next term. Finally, you repeat this process until you have proven that the inequality holds for all terms in the series.

4. What are some common applications of series inequality induction proof?

Series inequality induction proof is commonly used in various areas of mathematics, such as calculus, number theory, and combinatorics. It can also be applied in real-world scenarios, such as analyzing the growth of populations or the convergence of infinite series in physics and engineering.

5. Are there any limitations to series inequality induction proof?

Series inequality induction proof is a powerful tool for proving inequalities within a series, but it does have some limitations. It can only be used for discrete series, meaning that the terms in the series must be distinct and separated by a constant difference. It also requires a strong understanding of mathematical induction and may not be applicable in all mathematical scenarios.

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