Use the Limit Comparison Test to determine if the series is convergent: `sum_(n=3)^oo` n^2 -1/n^4 +2

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oldnick's profile pic

oldnick | (Level 1) Valedictorian

Posted on

`(n^2-1)/(n^4+2)<= (n^2-1)/(n^4-2n^2+1+1)<=(n^2-1)/((n^2-1)^2+1)<=(n^2-1)/(n^2-1)^2` `<=1/(n^2-1)`

The series :

 

`sum_(n=3)^oo 1/(n^2-1)` 

converges, having same behaviour of the series :

`sum_(n=3)^oo 1/n^2`

On the other side we saw:

`0<=sum_(n=3)^oo (n^2-1)/(n^4+2) <= sum_(n=3)^oo 1/(n^2-1)`  

So that  series:

`sum_(n=3)^oo (n^2-1)/(n^4+2) `

converges too.

 

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

What I understand question may be

`sum_(n=3)^oo(n^2-1)/(n^4+2)`

`` Let define `a_n=(n^2(1-1/n^2))/(n^4(1+2/n^4))`

`` `=(1-1/n^2)/(n^2(1+2/n^4))`

`a_n<1/n^2=b_n`

`sum_(n=3)^ooa_n<sum_(n=3)^oob_n`

By `sum_(n=3)^oo(1/n^p)`  ,comparision test it  converges if p>1.

`sum_(n=3)^oob_n`  converges ,so `sum_(n=3)^oo a_n`  will converge.

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