# please determine whether the series converges or diverges, and which test is appropriate to reach this determination: `sum_(n=1)^oo(2n+5)/(n^3-2)`

*print*Print*list*Cite

To determine if the series` sum_(n=1)^oo (2n+5)/(n^3 - 2)` converges or diverges, use the limit comparison test.

For two series `sum_(n=1)^oo a_n` and `sum_(n=1)^oo b_n` if `lim_(n->oo) a_n/b_n = L` and L is finite and positive either both the series converge or both of them diverge. For this test, the series `sum_(n=1)^oo 1/n^2` which is convergent and `sum_(n=1)^oo 1/n` which is divergent are being used.

`lim_(n->oo) (1/n^2)/((2n+5)/(n^3 - 2))`

= `lim_(n->oo) (n^3 - 2)/(2n^3+5n^2)`

= `lim_(n->oo) (n^3 + 2.5*n^2 - 2.5*n^2 - 2)/(2n^3+5n^2)`

=`lim_(n->oo) (n^3 + 2.5*n^2)/(2n^3+5n^2) - (2.5*n^2 + 2)/(2n^3+5n^2) `

= `lim_(n->oo) 1/2 - 2.5n^2/(2n^3+5n^2) - 2/(2n^3+5n^2)`

= `lim_(n->oo) 1/2 - 2.5/(2n+5) - 2/(2n^3+5n^2)`

As n tends to `oo` , `2.5/(2n+5)` and `2/(2n^3+5n^2)` tend to 0.

= `1/2`

As 1/2 is finite and positive, both the series `sum_(n=1)^oo 1/n^2` and `sum_(n=1)^oo (2n+5)/(n^3 - 1)` are convergent. There is no requirement to test with the other series that had been chosen earlier.

**The seriesĀ ` sum_(n=1)^oo (2n+5)/(n^3 - 2)` is convergent.**