I am self studying Understanding Analysis, and I was trying to do the following problem:
Using only the limit definition, prove that the sequence $\frac{n^3 + n}{2n^3+1}$ con- verges.
I know the sequence should converge to 1/2 (from calculus), but I'm lost on how to prove it. I am trying to do something similar this answer.
Since I know the sequence converges to 1/2, I tried: $$\lvert \frac{n^3 + n}{2n^3+1} - \frac{1}{2}\rvert < \epsilon$$ then $$\lvert \frac{2n-1}{4n^3+2} \rvert < \epsilon$$ $$\lvert \frac{2n-1}{4n^3+2} \rvert < \lvert \frac{2n-1}{4n^2-1} \rvert = \frac{1}{2n+1} < \epsilon$$ $$n > (\frac{1}{2})(\frac{1}{\epsilon} - 1)$$
So if we choose $N > (\frac{1}{2})(\frac{1}{\epsilon} - 1)$, for any n>N, we have: $$\lvert \frac{2n-1}{4n^3+2} \rvert < \frac{1}{2n+1} < \epsilon$$
I'm pretty sure the solution is wrong, since we should be saying the fraction is smaller than $\lvert \frac{2n-1}{4n^3+2}\rvert$ for the inequality to hold.