which of these series converge?

Question

sum of (1 / pi)^n? sum of 1/n? sum of (-1)^n / square root of n? from when n goes from 1 to infinity

Answer 1

The ratio test can be used ineach case.
1) n+1 th term = 1/π^(n+1)
nth term = 1/π^n

ratio = 1/π < 1
Therefore it converges

2) 1/n does not converge. Just know that.

3) n+1 th term = (-1)^(n+1) / sqrt(n+1)
nth term = (-1)^n / sqrt(n)
ratio = -sqrt [n / n+1]

As n goes to inf, ratio --> -1
So for the larger values of n, the successive terms cancel each other, thus the series will converge.

Answer 2

The first series will converge by the geometric series test because 1/pi<1.

The second series will diverge by the p-test, because p is equal to 1.

The third series will converge by the alternating series test.

Answer 3

a) sum = 1/(1-pi), so series does converge
b) well known not to converge
c) write sum as sum(n even) + sum (n odd) =
sum(1/sqrt(2m), m=1 to inf) -sum(1/(2m+1), m=1 to inf) +1
= sum((sqrt(2m+1) -sqrt(2m))/(sqrt(2m+1)sqrt(2m) )+1

Now look at the behavior as m -> inf. The top goes to sqrt(2m) x (1+1/2m) - sqrt(2m) =1/sqrt(2m). The bottom goes to 2m. Therefore the series goes as m^(-3/2), which converges.

Answer 4

I only would like to point out that ratio test is NOT applicable to alternate series