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Does this converge or diverge?

$$ \sum\limits_{n=1}^\infty (a_{n} = \frac{1}{2\sqrt{n} + \sqrt[3]{n}}) $$

The answer is: diverges by limit comparison to $\sum (b_{n} = \frac{1}{\sqrt{n}})$

If I look at $\lim_{n \to \infty}\frac{a_{n}}{b_{n}}$ I get

$$ \frac{\sqrt{n}}{2\sqrt{n} + \sqrt[3]{n}} = \frac{n^\frac{1}{2}} {2n^{\frac{1}{2}} + n^{\frac{1}{2}}n^\frac{2}{3}} = \frac{1}{2 + n^\frac{2}{3}} = \frac{0}{0 + 1} = 0 $$ as $n\to\infty$

I would expect it to be some $c > 0$ or $\infty$

Also, because $a_{n} < b_{n}$ I don't think I can use the comparison test. I'm pretty sure I am missing something but not sure what.

Thanks

Update:

My algebra was wrong, how is this for finding the limit?

$$ \frac{\sqrt{n}}{2\sqrt{n} + \sqrt[3]{n}} = \frac{1} {2 + \frac{1}{n^\frac{1}{6}}} = \frac{1} {\frac{2\sqrt[6]{n} + 1}{\sqrt[6]{n}}} = \frac{\sqrt[6]{n}}{2\sqrt[6]{n} + 1} = \frac{1}{2 + 0} = \frac{1}{2} $$ as $n \to\infty$

And since $\frac{1}{2} > 0$ and $b_{n}$ diverges, $a_{n}$ diverges

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You can use a straight comparison test with (one-third of) the harmonic series: $\frac1{2\sqrt{n}+\sqrt[3]{n}}\ge\frac1{3n}$. –  Brian M. Scott Oct 5 '12 at 10:11

2 Answers 2

up vote 0 down vote accepted

$$\frac{\sqrt{n}}{2\sqrt{n}+\sqrt[3]{n}}=\frac{\sqrt{n}}{\sqrt{n}\left(2+\frac{\sqrt[3]{n}}{\sqrt{n}}\right)}=\frac{1}{2+\frac{\sqrt[3]{n}}{\sqrt{n}}}\to\frac12,\text{ as $n\to\infty$}$$ because $$\frac{\sqrt[3]{n}}{\sqrt{n}} = n^{1/3}/n^{1/2} = n^{1/3-1/2}=n^{-1/6}$$


An other argument goes like this $$\sqrt[3]{n}\leq\sqrt{n},$$ so that $$2\sqrt{n}+\sqrt[3]{n}\leq3\sqrt{n} $$ and then $$\frac{\sqrt{n}}{2\sqrt{n}+\sqrt[3]{n}}\geq\frac{\sqrt{n}}{3\sqrt{n}}>1/5$$

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thanks very much! –  dizzy Oct 5 '12 at 11:17

There seems to be a mistake in the algebra -- namely, you took $\sqrt[3]{n} = n^{1/2} n^{2/3}$. But the right-hand side of this is equal to $n^{1/2 + 2/3} = n^{3/6 + 4/6} = n^{7/6} \ne \sqrt[3]{n}$.

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argh right you are. It seems I am not able to find that limit –  dizzy Oct 5 '12 at 10:07

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