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+...#?

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Answer 1 · 2018-06-08T14:11:16

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

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

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#