# Convergence of $\sum_{n=0}^{\infty} \left(\sqrt[3]{n^3+1} - n\right)$

I have encountered the following problem:

Determine whether $$\sum \limits_{n=0}^{\infty} \left(\sqrt[3]{n^3+1} - n\right)$$ converges or diverges.

What I have tried so far:

Assume that $a_n = \sqrt[3]{n^3+1} - n$.

• $n$th term test: $\lim \limits_{x\to\infty} a_n = 0$ $\Rightarrow$ inconclusive.
• $p$-series test: can't be used
• geometric series test: can't be used
• alternating series test: not alternating, can't use
• telescoping series: does not look like it (after rewriting first few terms)
• integral test: integrating the $a_n$ is outside the scope of my course (checked result with wolfram)
• ratio test: simplification of the limit is not possible due to number of expanding terms
• comparison / limit comparison test: no idea what should I compare it to

I would be grateful for any help.

• Recall that $$\forall a,b\in \mathbb R\left(a^3-b^3=(a-b)\left(a^2+ab+b^2\right)\right).$$ Find $$\forall n\in \mathbb N\left(\root 3\of {n^3+1}-n=\dfrac{1}{\root 3\of {n^3+1}^2+n\root 3\of {n^3+1}+n^2}\right)$$ and compare. – Git Gud May 18 '15 at 10:18

Hint We can apply a cubic analogue of the technique usually called multiplying by the conjugate, which itself is probably familiar from working with limits involving square roots. In this case, write $$u = \sqrt[3]{n^3 + 1},$$ so that $$u^3 - n^3 = (n^3 + 1) - (n^3) = 1.$$ On the other hand, factoring gives (in analogy to the difference of perfect squares factorization $a^2 - b^2 = (a - b)(a + b)$ used in the square root case case) $$u^3 - n^3 = (u - n) (u^2 + un + n^2).$$

Explicitly, multiplying gives that we can write our sum as \begin{align} \sum_{n = 0}^{\infty} (u - n) &= \sum_{n = 0}^{\infty} (u - n) \cdot \frac{u^2 + un + n^2}{u^2 + un + n^2} \\ &= \sum_{n = 0}^{\infty} \frac{u^3 - n^3}{u^2 + un + n^2} \\ &= \sum_{n = 0}^{\infty} \frac{1}{u^2 + un + n^2} . \end{align} Now, $u > n$, and so the summand satisfies $$\frac{1}{u^2 + un + n^2} < \frac{1}{3 n^2}.$$ Hence, by the comparison test, the given series is bounded by $$\frac{1}{3} \sum_{n = 0}^{\infty} \frac{1}{n^2},$$ which converges by, e.g., the $p$-test.

Hint.

Note

$\sqrt[3]{n^3+1} = n\sqrt[3]{1+1/n^3}$, and then $\sqrt[3]{1+1/n^3} = 1 + \frac{1}{3n^3} + o(\frac{1}{n^3})$.

Next, I belive, limit comparison test will work.

For any $x\in[0,1]$, the inequality: $$(1+x)^{\frac{1}{3}}\leq 1+\frac{x}{3} \tag{1}$$ holds by concavity or just by considering the cube of both terms. It follows that: $$0\leq \sqrt[3]{1+n^3}-n = n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\leq\frac{1}{3n^2}\tag{2}$$ hence the given series is convergent by the p-test or by the inequality: $$0\leq \sum_{n=1}^{N}\frac{1}{n^2} \leq 1+\sum_{n=1}^{N-1}\frac{1}{n(n+1)}=2-\frac{1}{N}\leq 2.\tag{3}$$

\begin{align*} \sqrt[3]{n^3+1}-n&=\frac{\left(\sqrt[3]{n^3+1}-n\right)\left(\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)}+n^2 \right )}{\left(\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)}+n^2 \right )}\\ &=\frac{1}{\left(\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)}+n^2 \right )}\\ &\le\frac{1}{3n^2}, \end{align*} so by direct comparison, your series converges.