# 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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`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"`

``

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.