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

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oldnick's profile pic

oldnick | (Level 1) Valedictorian

Posted on

  `sum_(n=3)^oo (n^3-1)/(n^4+2)`

We have to work on the general term  `a_n` .

We can note that if whe choice for comaprision a ratio `b_n ` with denominator greater tha those of  `a_n`  we get that `a_n >=b_n` 

For this we have: 

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

since : 

`n^3-1=(n-1)(n^2+n+1)` 

we get: 

`a_n>= (n-1)/(n^2+n+1)`  `>=(n-1)/(n+1)^2` `>=(n+1-2)/(n+1)^2` `=1/(n+1) -2/(n+1)^2`

Now the series :

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

si composed by the term    `1/(n+1)`   and  `2/(n+1)^2`

That is:

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

 

The second series at right side  converges having same behaviour of  seires `1/n^2`  not affecting behaviour of first series, since it is of infinitesimal lower, the series `1/(n+1)`  tha has same behaviour of the series `1/n`  that diverges.

Thus :

   `sum_(n=3)^oo b_n`  

diverges, being  for everi n: `b_n<= a_n`  then, series

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

`"diverges"`

``

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

What i understand again typographical error

`sum_(n=3)^ooa_n`  ,where `a_n=(n^3-1)/(n^4+2)`

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

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

Let define `sum_(n=3)^oob_n,where`

`b_n=1/n`

`a_n<b_n `      ,  for all n .

But series `sum_(n=n)^oob_n`  diverges  by Cauchy's General principle of convergenc.

`therefore `  series `sum_(n=3)^ooa_n`  will diverge.

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