# convergence tests for series

• May 27th 2009, 08:54 AM
Sniperpro
convergence tests for series
what would be an appropriate test (out of ratio, intergral, limit comparison) to decide if the series

sum of(1 to infinity)[k^(1/2)/ $(3*k^3+1)$]

converges or diverges?
• May 27th 2009, 09:18 AM
apcalculus
Quote:

Originally Posted by Sniperpro
what would be an appropriate test (out of ratio, intergral, limit comparison) to decide if the series

sum of(1 to infinity)[k^(1/2)/ $(3*k^3+1)$]

converges or diverges?

A (limit) comparison test would work best... certainly would not try the integral test.
• May 27th 2009, 09:31 AM
Amer
you can use ratio test I prefer it

$lim_{k\rightarrow\infty}\frac{\sqrt{k+1}(3(k)^3)}{ ( \sqrt{k}(3(k+1)^3+1)}=lim_{k\rightarrow\infty}\fra c{\sqrt{k+1}(3k^3)}{( \sqrt{k}(3k^3+3k^2+3k+1+1)}$
$=lim_{k\rightarrow\infty}\frac{\frac{\sqrt{k+1}}{k ^3}(3)}{\frac{\sqrt{k}}{k^3}(3+\frac{3}{k})+\frac{ 3}{k^2}+\frac{2}{k^3})}$
$=lim_{k\rightarrow\infty}\frac{\frac{\sqrt{k+1}}{k ^3}(3)}{\frac{\sqrt{k}}{k^3}(3+\frac{3}{k})+\frac{ 3}{k^2}+\frac{2}{k^3})}=1$

$the.... test... failed ...try.. comparison... test$
• May 27th 2009, 09:40 AM
Amer
$\sum{\frac{\sqrt{k}}{3k^3+1}}<$ $\sum{\frac{k}{3k^3}}$
• May 27th 2009, 12:32 PM
Sniperpro
so would it be ok use k^(1/2)/k^3 as Bn in the limit comparison test?
• May 27th 2009, 02:52 PM
Amer
No i think it will be the same take
$\frac{k}{3k^3}>\frac{\sqrt{k}}{3k^3+1}$

$and..\frac{k}{3k^3}=\frac{1}{3k^2}$

$\sum{\frac{1}{3k^2}} .converge .....by...integral$