How can I show that the series [(-1)^n]/n^(1/3) converges?

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- May 19th 2008, 02:44 PMjules027Series convergence
How can I show that the series [(-1)^n]/n^(1/3) converges?

- May 19th 2008, 02:57 PMMathstud28
$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}$

By the alternating series test since $\displaystyle \forall{n}>0,\frac{1}{\sqrt[3]{n}}>\frac{1}{\sqrt[3]{n+1}}$

so for an arbitrary value we have showed that for every value greater than that value that

$\displaystyle a_{n+1}<a_n$

Now we must show that $\displaystyle \lim_{n\to\infty}\bigg|\frac{(-1)^n}{\sqrt[3]{n}}\bigg|=\lim_{n\to\infty}\frac{1}{\sqrt[3]{n}}=0$

So we have also established that $\displaystyle \lim_{n\to\infty}|a_n|=0$

So this series conditionally converges

but it does not absolutely converge since

$\displaystyle \sum_{n=1}^{\infty}\bigg|\frac{(-1)^n}{\sqrt[3]{n}}\bigg|=\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}}$

diverges by the integral test

since it meets all its requirements and

$\displaystyle \int_1^{\infty}\frac{1}{\sqrt[3]{n}}dn=\infty$ - May 19th 2008, 03:42 PMjules027
That makes sense. This was from a bigger problem: "Give an example of two convergent series

**an**and**bn**such that the series(**an*****bn**) diverges. So would**an**being [(-1)^n]/n^(1/3) and**bn**being [(-1)^n]/n^(1/2) work for this problem? - May 19th 2008, 03:54 PMMathstud28
Assuming you mean $\displaystyle \sum_{n=1}^{\infty}\bigg[\frac{(-1)^n}{\sqrt{n}}\cdot\frac{(-1)^n}{\sqrt[3]{n}}\bigg]$ and not $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}\cdot\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}$

then yes you are correct the multiplication of both would yield $\displaystyle \frac{1}{n^{\frac{5}{6}}}$

which diverges by the integral test