# Series inequality induction proof

• Aristarchus_

#### Aristarchus_

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...

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.

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?

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 $$1+2^{2} + 3^{2} + \dots +n^{2} > \frac {1}{3} n^{3}$$ I would use $$1 + 2^2 + 3^2 + \dots + n^2 = \tfrac16n(n+1)(2n+1).$$ 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.

SammyS
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?