I need to determine if the following limit exists:
$$\sum_{n=1}^\infty (\sqrt[3]{n+1}-\sqrt[3]{n-1})^\alpha$$
Where alpha is some real number (The answer might depend on the choice of alpha).
This is the only question in my homework I was not able to do, I tried to define $a_n=(\sqrt[3]{n+1}-\sqrt[3]{n-1})^\alpha$ and look if something interesting happens with $\frac{a_{n+1}}{a_n}$ or with $\sum2^n \cdot a_{2^n}$ but unfortunately it did not work. Thank you for your help, and if you answer me please try to make your answer as basic as possible, because I'm not advanced yet so I don't know much (it should be basic, as this is a homework question at calculus 1 course). Thank you!


The key issue here is that you need to exploit some sort of cancellation between the two halves of the summands; the trivial estimate that the summands are just about bounded by $(2n^{1/3})^{\alpha}$ when $\alpha > 0$ isn't good enough.

A standard trick to study these is to use the binomial theorem to give a sharp approximation of the power function. To wit,

$$(n + 1)^{1/3} = n^{1/3}\left(1 + \frac 1 n\right)^{1/3} = n^{1/3} \left(1 + \frac 1 3 \frac 1 n + O(n^{-2})\right) \approx n^{1/3} + \frac 1 3 n^{-2/3}$$

Using a similar approximation for the other half of the summand, the question you really need to answer is for which $\alpha$ the series

$$\sum_{n = 1}^{\infty} \left(n^{-2/3}\right)^{\alpha}$$


  • $\begingroup$ That's a great answer! It's totally beside the point, but I need to ask: what does "to wite" mean? $\endgroup$ – rafa11111 Dec 2 '18 at 22:20
  • $\begingroup$ @rafa11111 Thanks! It was a typo of to wit, a slightly archaic way to say "namely." $\endgroup$ – user296602 Dec 2 '18 at 22:21
  • $\begingroup$ Hey, thanks for your answer! can you just explain to me what does that O(n^-2) mean? never saw it before. thanks! $\endgroup$ – Omer Dec 2 '18 at 22:22
  • 1
    $\begingroup$ @Omer It's "big O notation," and it just means that there is an error term which is no larger than (a multiple of) $n^{-2},$ which is negligible in this context. $\endgroup$ – user296602 Dec 2 '18 at 22:23
  • $\begingroup$ That's really a too complicated method for an asker that is asking for "please try to make your answer as basic as possible, because I'm not advanced yet so I don't know much". You should read carefully the OP before to answer to it. Refer to my answer based only on algebraic methods. $\endgroup$ – user Dec 2 '18 at 22:47

As a simpler method recall that

$$A^3-B^3=(A-B)(A^2+AB+B^2) \implies A-B=\frac{A^3-B^3}{A^2+AB+B^2}$$


$$\sqrt[3]{n+1}-\sqrt[3]{n-1}=\frac{2}{\sqrt[3]{(n+1)^2}+\sqrt[3]{n^2-1}+\sqrt[3]{(n-1)^2}} \sim\frac2{3n^\frac23}$$

then the given series converges by limit comparison test with $\sum \left(\frac1{n^{2/3}}\right)^\alpha$ for $\frac23\alpha >1$ that is $\alpha>\frac32$.


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