<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] 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
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