# Thread: Determine whether the series converges or diverges.

1. ## Determine whether the series converges or diverges.

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

Will use comparison test:

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

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

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

3. ## Re: Determine whether the series converges or diverges.

Thanks, I had to change the latex

4. ## Re: Determine whether the series converges or diverges.

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

Will use comparison test:

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

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

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

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

5. ## Re: Determine whether the series converges or diverges.

Originally Posted by Plato
Just note that $\displaystyle \frac{\sqrt{n^4 + 5n + 10}}{(3n+2)^2(5n+1)}\le \frac{4n^2}{(3n+2)^2(5n+1)}$
How did you get the 4 in the numerator?

6. ## Re: Determine whether the series converges or diverges.

Originally Posted by moonman
How did you get the 4 in the numerator?

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

7. ## Re: Determine whether the series converges or diverges.

Originally Posted by Plato
$\displaystyle \sqrt{n^4 + 5n + 10}\le\sqrt{n^4 + 5n^4 + 10n^4}=4n^2$
How? I'm sorry I don't understand how you get 4n^2

8. ## Re: Determine whether the series converges or diverges.

Originally Posted by moonman
How? I'm sorry I don't understand how you get 4n^2
$\displaystyle 4n^4+5n^4+10n^4=16n^4$

What is the square root of that?

9. ## Re: Determine whether the series converges or diverges.

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