converges or diverges

**simsima_1** · April 25th 2008, 11:05 AM

$<br /> \sum_1^{\infty} \frac{1}{\sqrt{{k^4}+3k}}<br />$

what test would u use for this? i'm thinking its prolly a telescoping series but i have no idea what to do.

**Plato** · April 25th 2008, 11:09 AM

$\frac{1}{{\sqrt {k^4 + 3k} }} < \frac{1}{{k^2 }}$

**simsima_1** · April 25th 2008, 11:17 AM

thats not true though...k^4 i < (k^4+3K)^1/2 .... does this prove divergance?

**Moo** · April 25th 2008, 11:21 AM

Hello,

$k^4+3k>k^4$

$\sqrt{k^4+3k}>\sqrt{k^4}=k^2$ because $f:x\mapsto \sqrt{x}$ is an increasing function.

$\frac{1}{\sqrt{k^4+3k}}<\frac{1}{k^2}$ because $g:x\mapsto \frac{1}{x}$ is a decreasing function.

And we know that $\sum \frac{1}{k^a}$ converges if $a>1$

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