How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given 1+3/4+9/16+...+(3/4)^n+...?

1 Answer
Jun 8, 2018

sum_(k=0)^n (3/4)^k = 4-3(3/4)^(n-1)

sum_(k=0)^oo (3/4)^k = 4

Explanation:

This is a geometric series of ratio q = 3/4.

Consider the series:

sum_(k=0)^oo q^k

and its partial sum:

s_n = sum_(k=0)^n q^k = 1+q+q^2+...+q^n = (q^(n+1)-1)/(q-1)

Then, if abs q < 1:

lim_(n->oo) s_n = lim_(n->oo) (q^(n+1)-1)/(q-1) = 1/(1-q)

For q=3/4:

s_n = ((3/4)^n-1)/(3/4-1) = (3^n-4^n)/(4^n(-1/4))=(4^n-3^n)/4^(n-1) = 4-3(3/4)^(n-1)

and:

sum_(k=0)^oo (3/4)^k = 4