determine whether the series is convergent or divergent.?

Question

a) summation notation symbol ( n=1 to infinity) of 1/(2n+1)^3
b)summation notation symbol ( n=1 to infinity) of (3+sin(n))/(n^2)
c) summation notation symbol(n=1 to infinity) of n/((sqrt(3n+2))

Answer 1

Nestor is right, mostly (but not always) for the right reasons:

a) 1/(2n+1)^3 ~ (1/8)*(1/n^3) ; and this converges, as Nestor said, because 3 > 1, and so converges by the integral test.

b) (3 + sin(n))/n^2 = 3/n^2 + sin(n)/n^2.
The 3/n^2 surely converges.
The sin(n)/^2 converges because it converges absolutely (i.e., the sum of the absolute value converges).
So the sum converges.

c) a_n = n/sqrt(3n+2) => n/sqrt(3n) = (1/sqrt(3))*sqrt(n)
[NOT, as Nestor thought, 1/sqrt(n)]
So a_n => infinity, and cannot give a convergent sum!

Answer 2

the serie of 1/ n^a is convergent if a > 1

the first is like the serie of 1/ n³ then convergent

the second is between the series of 2/n² and 4/n², then convergent

the third is like 1 /sqrt(n) or 1 / n^0.5 then divergent



All right Nealjking ! I have not seen the n over the square root in the third question : the serie is
" more " divergent so !