Who can Solve THIS ???

Question

a) find the values of p for which the series is convergent

Sum(from 1 to infinity) (ln(n)/n^p)
b) Determine whether the series is convergent or divergent

Sum(from 0 to infinity) (1+sin(n))/10^n

Sum(from 1 to infinity) (n+5)/(n^7+n^2)^(1/3)

Answer 1

a) ln(n)/n^p os convergent for p>1 (strictly)
To prove it I suppose you know that 1/n^p for p> 1 i convergent
and that ln(n)/n^d with d>0 has lim 0(you can prove that with L´Hòpital)

so if p> 1 p= 1+d/2 +d/2 so
ln(n)/n^p = ln(n)/n^d/2 * 1/n^(1+d/2) < 1/n^(1+d/2) as
ln(n) n^d/2 having lim 0 is <1
so ln(n) n^p < 1/n^(!+d/2) which is convergent and so is also convergent

(1+sin n)/10^n <= 2/10^n and this is a geometric series with r=1/10<1 convergent

The last is of the same class as 1/n^(7/3-1) = 1/n^4/3 which is convergent.
(Divide your term by 1/n^4/3 and you´ll that the limit is a number non zero)