Using the definition of divergence to infinity, proove n^3 + n diverges to infinity

[hexed sonnet]

I am given a question to use the definition of convergence to prove that n^3+n diverges to infinity.

Attempt at solution:

Recall that a sequence a_n diverges to positive infinity if given M > 0 there exists an N in the natural numbers such that whenever n > N we have a_n > M.

Solve n^3 + n > M

n(n^2 + 1) > M

. . . then I get stuck

[lime rock]

So far you’re doing good. You now need to pick a value for N now such that for any n > N the inequality holds. You should pick N to involve the variable M

If it is not immediately obvious try choosing the most simple choice you can think of and see if it works. If not think about why it fails and adjust it so it works

[hexed sonnet]

I chose n > sqrt M-1

I chose an expression smaller but bigger than M and arrived at n > sqrt M-1

So I choose N to be sqrt M-1

@lime rock

[lime rock]

Alright that works. Now you know N(N^2-1) > M, and in general if n>N then n(n^2-1)>N(N^2-1). Now you just put everything together to finish the proof

BTW if one choice of N works a larger choice of N also works. In this case you conjecture sqrt M-1 works. If you are correct about this (you are) then you also know N=M works. Using this will make the proof nicer. My point is if you believe one value works, try doing the proof with a slightly bigger but simpler value. In this case N=M is much simpler than N=sqrt M-1

[hexed sonnet]

Got it I'll take that advice and use that to develop my understanding.

Thank you very much!