Find x for which $\sum\left[(n^3+1)^\frac13-n\right]x^n$ converges.

Question

For what values of $x$ the infinite series

$\sum\left[(n^3+1)^\frac13-n\right]x^n$ converges?

Please help me out.

Using ratio test

$\begin{aligned}&\lim_{n\to \infty}\frac{(n^3+1)^{1/3}-n}{\left[\left((n+1)^3+1\right)^{1/3}-(n+1)\right]x}&>1\\\implies&\lim_{n\to \infty}\frac{n[\left(1+\frac1{n^3}\right)^{1/3}-1]}{n\left[\left(\left(1+\frac1n\right)^3+\frac1{n^3}\right)^{1/3}-\left(1+\frac1n\right)\right]x}&>1\\\implies&\lim_{n\to \infty}\frac{\left[\left(1+\frac1{3n^3}\right)-1\right]}{\left[\left(\left(1+\frac1n\right)^3+\frac1{n^3}\right)^{1/3}-(1+\frac1n)\right]x}&>1\end{aligned}$

But I can't understand how will I approximate the denominator $\left(\left(1+\frac1n\right)^3+\frac1{n^3}\right)^{1/3}$

Please help me out.

Answer 1

Recall the difference of two cubes factorisation: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2).$$ Now consider the above with $a = (n^3 + 1)^{1/3}$ and $b = n$. Then, $$((n^3 + 1)^{1/3} - n)((n^3 + 1)^{2/3} + n(n^3 + 1)^{1/3} + n^2) = n^3 + 1 - n^3 = 1.$$ Rearranging, $$(n^3 + 1)^{1/3} - n = \frac{1}{(n^3 + 1)^{2/3} + n(n^3 + 1)^{1/3} + n^2}.$$ Try applying the ratio test again, this time on the same sum: $$\sum \frac{1}{(n^3 + 1)^{2/3} + n(n^3 + 1)^{1/3} + n^2}x^n.$$ It should help.

Answer 2

Hint: We know power series are absolutely convergent in their radius of convergence, hence

$$\left|\sum_{n=1}^\infty\left[(n^3+1)^{1/3}-n\right]x^n\right|\leq\sum_{n=1}^\infty 3nx^n,$$the latter having a radius of convergence $1$ (i.e., the former cannot have a radius of convergence exceeding $1$). I leave it to you to show the radius of convergence equals $1$ (but this would be my general approach).

Answer 3

More hints: for $x>0$, $ \sum_{n\ge1} x^n n^k $ converges for all $x<1$ and any $k$. Similarly, it diverges for all $x>1$ and any $k$. So the only thing you need to check is when $x=1$.