# Find the smallest positive integer

#### youngstudent16

1. The problem statement, all variables and given/known data
For a positive integer $n$, let

$a_n=\frac{1}{n} \sqrt[3]{n^{3}+n^{2}-n-1}$

Find the smallest positive integer $k \geq2$ such that $a_2a_3\cdots a_k>4$
2. Relevant equations

The restrictions are the only relevant thing I can think of
3. The attempt at a solution

I have just tried plugging in numbers so far

When $n=2$ I got $\frac{3^{\frac{2}{3}}}{2}$
When $n=3$ I got $\frac{2 \times 2^{\frac{2}{3}}}{3}$
When $n=4$ I got $\frac{1}{4} 3^{\frac{1}{3}} \hspace{1mm} 5^{\frac{2}{3}}$

Now this is growing really slowly so this is obviously not the correct approach.

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#### Mark44

Mentor
1. The problem statement, all variables and given/known data
For a positive integer $n$, let

$a_n=\frac{1}{n} \sqrt[3]{n^{3}+n^{2}-n-1}$

Find the smallest positive integer $k \geq2$ such that $a_2a_3\cdots a_k>4$
2. Relevant equations

The restrictions are the only relevant thing I can think of
3. The attempt at a solution

I have just tried plugging in numbers so far

When $n=2$ I got $\frac{3^{\frac{2}{3}}}{2}$
When $n=3$ I got $\frac{2 \times 2^{\frac{2}{3}}}{3}$
When $n=4$ I got $\frac{1}{4} 3^{\frac{1}{3}} \hspace{1mm} 5^{\frac{2}{3}}$

Now this is growing really slowly so this is obviously not the correct approach.
I would advise you to just keep going. Sometimes, a brute force approach is the easiest way to go. For your results for n = 4, is the expression you show less than 4? Writing the result as a decimal approximation would be helpful.

#### Student100

Gold Member
1. The problem statement, all variables and given/known data
For a positive integer $n$, let

$a_n=\frac{1}{n} \sqrt[3]{n^{3}+n^{2}-n-1}$

Find the smallest positive integer $k \geq2$ such that $a_2a_3\cdots a_k>4$
2. Relevant equations

The restrictions are the only relevant thing I can think of
3. The attempt at a solution

I have just tried plugging in numbers so far

When $n=2$ I got $\frac{3^{\frac{2}{3}}}{2}$
When $n=3$ I got $\frac{2 \times 2^{\frac{2}{3}}}{3}$
When $n=4$ I got $\frac{1}{4} 3^{\frac{1}{3}} \hspace{1mm} 5^{\frac{2}{3}}$

Now this is growing really slowly so this is obviously not the correct approach.
Set up an inequality.

#### Student100

Gold Member
I would advise you to just keep going. Sometimes, a brute force approach is the easiest way to go. For your results for n = 4, is the expression you show less than 4? Writing the result as a decimal approximation would be helpful.
Never mind, I'm an idiot. :)

#### youngstudent16

Set up an inequality.
Ok trying to just work with the variables no numbers I'm getting this pattern

$\frac{\sqrt[3]{k-1)(k+1)^2}(k-1)}{k}$
Now I simplify and set up the ineqaulity
$\frac{\sqrt[3]{(2k^2)(k+1)^2}}{2k}>4$

Now what

#### youngstudent16

I would advise you to just keep going. Sometimes, a brute force approach is the easiest way to go. For your results for n = 4, is the expression you show less than 4? Writing the result as a decimal approximation would be helpful.
I tried putting them with decimal and using wolframalpha I went to 10 numbers and still it was tiny

#### Student100

Gold Member
Ok trying to just work with the variables no numbers I'm getting this pattern

$\frac{\sqrt[3]{k-1)(k+1)^2}(k-1)}{k}$
Now I simplify and set up the ineqaulity
$\frac{\sqrt[3]{(2k^2)(k+1)^2}}{2k}>4$

Now what
$a _2+a _3 + ... a _k$
Whats the sum of your brute force method? You are summing them correct?

#### youngstudent16

$a _2+a _3 + ... a _k$
Whats the sum of your brute force method? You are summing them correct?
Its multiplying not summing and yes I tried brute force now using wolframalpha for several numbers and its going up very slowly like 1.01 1..., 1..., 1... etc

#### Student100

Gold Member
Its multiplying not summing and yes I tried brute force now using wolframalpha for several numbers and its going up very slowly like 1.01 1..., 1..., 1... etc
Okay, well then as long as you set up your inequality right have you tried punching it into a calculator/wolfram and seeing if a solution exists? Picking various k's and working it out?

#### haruspex

Homework Helper
Gold Member
2018 Award
$\frac{\sqrt[3]{(2k^2)(k+1)^2}}{2k}>4$

Now what
You've done the hard work. Just cube both sides, multiply out and simplify.

#### Mark44

Mentor
As it turns out, brute force and direct calculation with paper and pencil don't work very well for this problem. You can, however, use brute force and a computer to find the answer. I put together an Excel spreadsheet that shows that when n is about 250, the product finally gets to 4.

The first two rows of my spreadsheet look like this:
Code:
 2 | 1/A1 * (A1^3 + A1^2 - A1 - 1)^(1/3) | =B1
=A1 + 1 | 1/A2 * (A2^3 + A2^2 - A2 - 1)^(1/3) | =B2 * C1
I just copied the second row (all three columns) down a bunch of rows.

#### haruspex

Homework Helper
Gold Member
2018 Award
As it turns out, brute force and direct calculation with paper and pencil don't work very well for this problem. You can, however, use brute force and a computer to find the answer. I put together an Excel spreadsheet that shows that when n is about 250, the product finally gets to 4.

The first two rows of my spreadsheet look like this:
Code:
 2 | 1/A1 * (A1^3 + A1^2 - A1 - 1)^(1/3) | =B1
=A1 + 1 | 1/A2 * (A2^3 + A2^2 - A2 - 1)^(1/3) | =B2 * C1
I just copied the second row (all three columns) down a bunch of rows.
It's not at all difficult algebraically. Look at where youngstudent got to in post #5.

#### andrewkirk

Homework Helper
Gold Member
Ok trying to just work with the variables no numbers I'm getting this pattern

$\frac{\sqrt[3]{k-1)(k+1)^2}(k-1)}{k}$
Now I simplify and set up the ineqaulity
$\frac{\sqrt[3]{(2k^2)(k+1)^2}}{2k}>4$

Now what
I'm not sure that's quite right.
Go back to the expression inside the cube root in $a_n$. Notice that it factorises to $(n^2-1)(n+1)=(n-1)(n+1)^2$.
Next, bring the $n$ in the denominator inside the cube root, by cubing it. That gives us
$${a_n}^3=\frac{(n-1)(n+1)^2}{n^3}$$

Hence the inequality you want to prove is:

$$4^3=64< \prod_{k=2}^n \frac{(k-1)(k+1)^2}{k^3}$$

Look carefully at the inside of the product. Notice how the power in the denominator is 3 and the numerator has the 1st power of the previous factor and the 2nd power of the next factor. Does that give you an idea about some really nice simplifying cancellation that is going to happen between adjacent factors?

Try writing out a few factors in a row like this and you'll get an idea of the cancellation, which will lead you towards a simple guessed expression for the product, in which everything cancels out except a few bits from the first and last factors. You can then use mathematical induction to prove that that expression is the correct one for each product.

Once you have that expression, solving the inequality is easy. It should be an inequality involving only polynomials in $n$, none with degree greater than 2, so it's just solving a quadratic.

Last edited:

#### haruspex

Homework Helper
Gold Member
2018 Award
I'm not sure that's quite right.
It's what I get. To get that, youngstudent has already done most of what you describe. Reducing it to a quadratic is all that's left.

#### youngstudent16

You've done the hard work. Just cube both sides, multiply out and simplify.
Ah perfect thank you I got 254 as the correct solution that was a lot of computation

#### andrewkirk

Homework Helper
Gold Member
It's what I get. To get that, youngstudent has already done most of what you describe. Reducing it to a quadratic is all that's left.
It's the first of the two formulas that I think is not right. The second is the same as mine, minus some cancelling ($\frac{(n+1)^2}{4n}<4^3$) but is not equivalent to the first. It's possible that the difference between the two is just an error in latex coding.
The purpose of my post was to indicate that one can do this problem deductively rather than just inductively.

"Find the smallest positive integer"

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