Find the convergence of $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ I want to find what the series $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ converges to exactly or show that it diverges.
By taking the partial sum of the series $S_N$ = $ \sum_{n=1}^{N} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ then $S_N = 2^\frac{1}{3} - 1 + 3^\frac{1}{3} - 2^\frac{1}{3} +4^\frac{1}{3}-3^\frac{1}{3} + ... + (N+1)^\frac{1}{3} - N^\frac{1}{3}$
And at the end I'm left with $S_N = -1 + (N+1)^\frac{1}{3}$ and $\lim_{N \to \infty} S_N = -1 + \infty= \infty $
So  $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3} = \infty$
Is this correct is my first question and my second question is does there exist any other method of finding what series to converge exactly?
Thank you in prior.

we are allowed to write each summand as we wish, so I will try the sum of $-n^{1/3} + (n+1)^{1/3}$ to get, up to $n = N,$
$$ \small -1 + 2^{1/3} - 2^{1/3} + 3^{1/3} - 3^{1/3} + 4^{1/3}  \cdots -(N-2)^{1/3}+(N-1)^{1/3} - (N-1)^{1/3} + N^{1/3} - N^{1/3} +(N+1)^{1/3}  $$
Just a visual thing, the pairs that cancel are next to each other this way
