# Series convergence

• May 19th 2008, 02:44 PM
jules027
Series convergence
How can I show that the series [(-1)^n]/n^(1/3) converges?
• May 19th 2008, 02:57 PM
Mathstud28
Quote:

Originally Posted by jules027
How can I show that the series [(-1)^n]/n^(1/3) converges?

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

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

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

$a_{n+1}

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

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

So this series conditionally converges

but it does not absolutely converge since

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

diverges by the integral test

since it meets all its requirements and

$\int_1^{\infty}\frac{1}{\sqrt[3]{n}}dn=\infty$
• May 19th 2008, 03:42 PM
jules027
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 PM
Mathstud28
Quote:

Originally Posted by jules027
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?

Assuming you mean $\sum_{n=1}^{\infty}\bigg[\frac{(-1)^n}{\sqrt{n}}\cdot\frac{(-1)^n}{\sqrt[3]{n}}\bigg]$ and not $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}\cdot\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}$

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

which diverges by the integral test