# Test the convergence of a series

To test the convergence of a series:

$$\sum\left[\sqrt[3]{n^3+1}-n\right]$$

My attempt:

Take out $n$ in common: $\displaystyle\sum\left[n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\right]$.

So this should be divergent.

But, the given answer says its convergent.

• "So this should be divergent" - why?
– M.M
Aug 20, 2014 at 20:22
• due to the "n". I took the $1/n^3$ to be zero. I should have used the taylor series to expand it. Aug 21, 2014 at 9:59

$$\sqrt[3]{n^3+1}-n=\frac{1}{\sqrt[3]{n^3+1}^2+n\sqrt[3]{n^3+1}+n^2}\le\frac{1}{3n^2}$$

• You can use $\leq$ instead of $\sim$. It makes for a more "solid" proof of convergence. Aug 20, 2014 at 11:41
• Nice suggestion! Aug 20, 2014 at 11:44

We have that $$\sqrt[3]{1+\frac{1}{n^3}} \sim_\infty 1+\frac{1}{3n^3}$$ by Taylor series.

So $$n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\sim_\infty\frac{1}{3n^2}.$$

Do you see why the series is convergent now? :)

• I'm affraid I don't follow. The notation $\sim_\infty$ indicates that the ratio of the two sequences tends to $1$ as $n\to\infty$, right? The limit of the sequence $\sqrt[3]{1+\frac1{n^3}}$ is $1$ so that we can write $\sqrt[3]{1+\frac1{n^3}}\sim_\infty x_n$ for any sequence $x_n$ that has $1$ as its limit. Could you provide more details? Aug 20, 2014 at 11:50
• Sure, good questions. In general the Taylor series gives us the approximation $$\big(1+\frac{1}{x}\big)^a =_\infty 1+\frac{a}{x}+o\big(\frac{1}{x}\big).$$ This approximation is frequently useful in evaluating limits. In our example we have expanded two terms of the Taylor series. Note that this was necessary to get a nonzero expression. Aug 20, 2014 at 12:20

$$\sqrt[3]{n^3+1}-n = \left({\sqrt[3]{n^3+1}-n)}\right) \frac{ \left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n + n^{2 } }\right)}{\left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n + n^{2 } }\right)} = \frac {1}{\left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n + n^{2 } }\right)} \le \frac{1}{(n^3 + 1)^{\frac 2 3} } \le \frac{1}{(n^3)^{\frac 2 3}} = \frac 1 {n^2}$$

And we know that $\sum \frac {1}{n^2}$ converges and by the Comparison Test our series converges too.

And a series will not diverge just because $n$ is a factor. What about $$\sum n \cdot \frac{1}{n!} \;\;\; ??$$

This is quite similar to the other answers, but slightly different derivation of the estimate: $$n^3+1 \le n^3+3 +\frac3{n^3}+\frac1{n^6}=\left(n+\frac1{n^2}\right)^3$$ $$\sqrt[3]{n^3+1} \le n+\frac1{n^2}$$ $$\sqrt[3]{n^3+1} -n \le \frac1{n^2}$$

Now we can use comparison test.

If you prefer to have a tighter estimate, you can use $\left(n+\frac1{3n^2}\right)^3$ instead.