Determine whether the series converges or diverges.

**moonman** · November 26th 2012, 12:18 PM

$\sum_{n=1}^{\infty}\frac{\sqrt{n^4 + 5n + 10}}{(3n+2)^2(5n+1)}$

Will use comparison test:

$\sum a_{n} = \sum_{n=1}^{\infty}\frac{\sqrt{n^4 + 5n + 10}}{(3n+2)^2(5n+1)}$

$\sum b_{n} = \frac{n^2}{30n^3}$

Is this correct series for $\sum b_{n}$ ?

**coolge** · November 26th 2012, 12:24 PM

Can't see your series.

**moonman** · November 26th 2012, 12:28 PM

Thanks, I had to change the latex

**Plato** · November 26th 2012, 12:39 PM

Originally Posted by moonman

$\sum_{n=1}^{\infty}\frac{\sqrt{n^4 + 5n + 10}}{(3n+2)^2(5n+1)}$

Will use comparison test:

$\sum a_{n} = \sum_{n=1}^{\infty}\frac{\sqrt{n^4 + 5n + 10}}{(3n+2)^2(5n+1)}$

$\sum b_{n} = \frac{n^2}{30n^3}$

Is this correct series for $\sum b_{n}$ ?

Just note that $\frac{\sqrt{n^4 + 5n + 10}}{(3n+2)^2(5n+1)}\le \frac{4n^2}{(3n+2)^2(5n+1)}$

**moonman** · November 26th 2012, 12:41 PM

Originally Posted by Plato

Just note that $\frac{\sqrt{n^4 + 5n + 10}}{(3n+2)^2(5n+1)}\le \frac{4n^2}{(3n+2)^2(5n+1)}$

How did you get the 4 in the numerator?

**Plato** · November 26th 2012, 01:10 PM

Originally Posted by moonman

How did you get the 4 in the numerator?

$\sqrt{n^4 + 5n + 10}\le\sqrt{n^4 + 5n^4 + 10n^4}=4n^2$

**moonman** · November 26th 2012, 01:25 PM

Originally Posted by Plato

$\sqrt{n^4 + 5n + 10}\le\sqrt{n^4 + 5n^4 + 10n^4}=4n^2$

How? I'm sorry I don't understand how you get 4n^2

**Plato** · November 26th 2012, 01:32 PM

Originally Posted by moonman

How? I'm sorry I don't understand how you get 4n^2

$4n^4+5n^4+10n^4=16n^4$

What is the square root of that?

**moonman** · November 26th 2012, 01:41 PM

Oh gosh! Thanks, I feel like an idiot now. I did not follow the order of operations.

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