<s>[INST] 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.
 [/INST] Another idea:
$$
1=(n+1)-n = ((n+1)^{1/3}-n^{1/3})((n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}).
$$
Hence, roughly speaking, $(n+1)^{1/3}-n^{1/3}$ grows like $n^{-2/3}$. Hence, asymptotically, I'd expect $\sum_n (n+1)^{1/3}-n^{1/3} \sim \sum_n n^{-2/3}$, which is well-known to be diverging.
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