who can SOLVE IT ???

Question

Determine whether the series is convergent or divergent

a) Sum(from 1 to infinity) (5/n^4+4/n^(3/2))

b) Sum(from 1 to infinity) (n*e^(-n))

Answer 1

a is definitely convergent. Just split up the sums into:
Sum(from 1 to infinity) ( 5 / n^4 ) +
Sum(from 1 to infinity) ( 4/n^(3/2) )

Both these sums are convergent by the ratio test, which says that if your formula is ( 1/n^x ) and x is greater than 1, than your summation is convergent.

b is also convergent.
You can use a couple tests to solve this one but I think I prefer the root test. Just make a limit and find out if it is less than 1:

lim (n->infinity) (n * e^-n)
lim (n->infinity) (n / e^n)

Now you can use hopital's rule on this limit which says that if the limit comes out to be "infinity / infinity" or it comes out to be "0/0" Then you can take the derivative of the top and the bottom and solve it. So you get:

lim (n->infinity) ( 1/ e^n )

( 1/ e^(infinity) ) = 0

0 is less than 1 so by the root test this series is convergent.

*Note: when you use hopital's rule your not taking the derivative of the whole formula, your taking the derivative of the top and bottom seperately and putting them back into the same function.

Answer 2

5/n^4+4/n^3/2 is convergent as it is <5/n^4 which is convergent(it is 1/n^p with p>1)
n/e^n( apply ratio test)

(n+1)e^n+1 *e^n/ (n) ==> 1/e<1 so the series is convergent