Determine if the series is converges or diverges, if converges give answer?

Question

summation notation symbol (n=3 to infinity) 1/(3^n)
summation notation symbol (n=6 to infinity) (6^n)/(7^n)
please give a detail explanation i really need the help i have a quiz on this tomorrow

Answer 1

the first is a geometric series with r=1/3 but missing the terms
1+1/3 +1/9
so the sum is (1/(1-1/3) -13/9= 1/18
The secon is also geometric with r=6/7 with missing terms
1+6/7+(6/7)^2++++(6/7)^5 = [(6/7)^6-1)/(-1/7)] while the sum of the geometric series complete is
1/(1-6/7)=7
The result is 7-[(1-(6/7)^6)]*7 = 7*(6/7)^6

Answer 2

basically what you do is to find the ratio
of the n+1 th element to the nth element.
Then you elavualate the limit of that ratio and check whether it's less than or greater than 1.

first question
n+1 th term is = 1/(3^(n+1))
n th term is = 1/(3^(n))

the ratio is
[1/(3^(n+1)) ] [1/(3^(n))] = 1/3 Note that since this a number the limit of this number is also 1/3. Since the ratio is smaller than 1, the series converges.

if it was something like 1/3n than the limit would be 0 since as n goes to infinity 1/3n would go 0.

Likewise the second question
6^n / 7^n = (6/7)^n
the n+1th term is (6/7)^n+1
the nth term is (6/7)^n

the ratio is [(6/7)^n+1] / [(6/7)^n] = 6/7
which is smaller than 1, so the series converges.