0
$\begingroup$

This question already has an answer here:

I recently saw this question about the convergence of $$\sum_{n=0}^{\infty} \sqrt[3]{n^3+1} - n$$

I tried finding the convergence of this series by factoring out an $n^3$ so that the sum became,

$$\sum_{n=0}^{\infty} n\sqrt[3]{1 + \frac{1}{n^3}}$$ but couldn't figure out any way to simplify and solve this. Is there a simpler or better way to solve this?

$\endgroup$

marked as duplicate by Martin R, Did calculus May 8 '18 at 13:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Hint: $a-b = \frac{a^3-b^3}{a^2+ab+b^2}$. $\endgroup$ – MisterRiemann May 8 '18 at 9:30
4
$\begingroup$

$$a_n= \sqrt[3]{n^3+1} - n=n\left(\sqrt[3]{1 + \frac{1}{n^3}}-1 \right)$$ For large $n$, use the generalized binomial theorem or Taylor series to get $$\sqrt[3]{1 + \frac{1}{n^3}}=1+\frac{1}{3 n^3}-\frac{1}{9 n^6}+O\left(\frac{1}{n^9}\right)$$ $$a_n=\frac{1}{3 n^2}-\frac{1}{9 n^5}+O\left(\frac{1}{n^8}\right)=\frac{1}{3 n^2}+O\left(\frac{1}{n^5}\right)$$

$\endgroup$
5
$\begingroup$

Yes: using the equality $x^3-y^3=(x-y)(x^2+xy+y^2)$. It follows from it that\begin{align}\sqrt[3]{n^3+1}-n&=\sqrt[3]{n^3+1}-\sqrt[3]{n^3}\\&=\frac1{\sqrt[3]{n^3+1}^2+\sqrt[3]{n^3+1}\sqrt[3]{n^3}+\sqrt[3]{n^3}^2}\\&=\frac1{(n^3+1)^{2/3}+n(n^3+1)^{1/3}+n^2}.\end{align}So, the denominator behaves as $3n^2$ and so you can apply the comparaison test, comparing your series with $\sum_{n=1}^\infty\frac1{n^2}$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.