# Convergence of a Polynomial Series [duplicate]

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?

## marked as duplicate by Martin R, Did calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 8 '18 at 13:39

• Hint: $a-b = \frac{a^3-b^3}{a^2+ab+b^2}$. – MisterRiemann May 8 '18 at 9:30
$$a_n= \sqrt[3]{n^3+1} - n=n\left(\sqrt[3]{1 + \frac{1}{n^3}}-1 \right)$$ For large $n$, use the generalized binomial theorem or Taylor series to get $$\sqrt[3]{1 + \frac{1}{n^3}}=1+\frac{1}{3 n^3}-\frac{1}{9 n^6}+O\left(\frac{1}{n^9}\right)$$ $$a_n=\frac{1}{3 n^2}-\frac{1}{9 n^5}+O\left(\frac{1}{n^8}\right)=\frac{1}{3 n^2}+O\left(\frac{1}{n^5}\right)$$
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 $\sum_{n=1}^\infty\frac1{n^2}$.