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.
 A: Your computation is right, up to this point:
$$S_N = 2^\frac{1}{3} - 1 + 3^\frac{1}{3} - 2^\frac{1}{3} +4^\frac{1}{3}-3^\frac{1}{3} + ... +N^{1/3}- (N-1)^{1/3}+ (N+1)^\frac{1}{3} - N^\frac{1}{3}$$
Note that once you cancel all terms you are left with
$$S_N=(N+1)^{\frac{1}{3}}-1$$
A: 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
A: 
While summation of the telescoping series is trivial and immediately shows divergence of the series, the OP has asked if there are alternative approaches.   Herein, we give two straightforward ways forward.


METHODOLOGY $1$:
Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ reveals
$$(n+1)^{1/3}-n^{1/3}=\frac{1}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}> \frac{1}{3(n+1)^{2/3}}>\frac{1}{3(n+1)}$$
Hence, we have
$$\sum_{n=1}^N \left((n+1)^{1/3}-n^{1/3} \right)>\frac13 \sum_{n=2}^{N+1}\frac1n$$ 
Inasmuch as the harmonic series diverges, the series of interest diverges also.

METHODOLOGY $2$:
Using $(n+1)^{1/3}-n^{1/3}=n^{1/3}\left(1+\frac1{3n}+O\left(\frac1{n^2}\right)\right)-n^{1/3}=\frac1{3n^{2/3}}+O\left(\frac1{n^{5/3}}\right)$
Inasmuch as the series $\sum_{n=1}^\infty \frac{1}{n^p}$ diverges for $p\le 1$, the series of interest diverges also.
A: 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.
