This is incorrect, this is one of the most commonly encountered errors. If is an infinite series and if then diverges. It says nothing of whethere it will converge/diverge if the limit is zero. For this one try comparing your series toinfinite

E n^2/(2n^3+1)

n=1

For this i used Divergence test. Divided all terms by n^3. Ended up with 0/2. This shows that it will converge because it was 0.

This is correct as long as you make sure it is known that you are using Leibniz's Criterion (Alternating series test)2nd problem.

Infinite

E (-1)^n (2/sqrt(n))

n=1

Bn= 2/sqrt(n) Lim n-> infinity 2/sqrt(n) = 0

Bn = 2/sqrt(n) > 2/sqrt(n+1) = B(n+1)

This shows that it converges

This is a correct application.3rd Problem.

Infinity

E n/2n+3

n=1

Used divergence test again.

divided all 3 terms by n

got 1/2

since not 0 it diverges

Yes since and this series does converge by Leibniz's Criterion. But note that this also coverges absolutely.4th problem

Infinity

E (-1)^n+1/((n^2))+3)

n=1

I did the same thing as number 2 and got the same answer.