comparison test??

Question

summation n/3^n+1 from 1 to infinity have to compare with wat??
i hav to use what method to determine it's converge or diverge?

Answer 1

n/(3^n+1) < n/(3^n) = 1/2^n
but the series 1/2^n converges [ geometrical = (1/2)/(1-(1/2))
so our series must converge

Answer 2

If that would strictly be a comparison test, then you would have to compare that with a SURE CONVERGENT SERIES which has values that are ALL LARGER THAN ANY VALUE OF THE series you are using. However, I would recommend using ratio test to determine if it converges or not.

That would be

lim (X[n]/X[n+1]) = something

If the "something" equals zero, then it converges, if not, then it doesn't, or its not sure.

Hope that helps.

Answer 3

n(3^n+1) is of the same class as n/(3^n) as the limit of their quotient is 1
to n/3^n apply the ratio test r= (n+1)/n 3^n/3^(n+1) =1/3(1+1/n)
which has limit 1/3<1 convergent