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

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Answer 1 · 2017-07-30T08:36:20

#S_n =sum_(k=0)^n (1/3)^k = 1/2 (3^n-1)/3^(n-1)#

#S= sum_(k=0)^oo (1/3)^k = 3/2#

This is a geometric series of ratio #q=1/3#:

#S= sum_(k=0)^oo (1/3)^k = sum_(k=0)^oo q^k#

Consider the #n#-th partial sum of the series:

#S_n = sum_(k=0)^(n-1) q^k = 1+q+q^2+...+q^(n-1)#

and note that:

#(1+q+q^2+...+q^(n-1))(1-q) = 1-cancelq+cancelq -cancel(q^2)+cancel(q^2) +...-q^n =1-q^n#

so that:

#S_n = (1-q^n)/(1-q) = (1-1/3^n)/(1-1/3) = 3/2(3^n-1)/3^n = 1/2 (3^n-1)/3^(n-1)#

As #lim_(n->oo) q^n = lim_(n->oo) (1/3)^n = 0# we have

#lim_(n->oo) S_n = (1-q^n)/(1-q) = 1/(1-q) = 1/(1-1/3) = 3/2#

And in general we can see that a geometric series converges when:

#lim_(n->oo) q^n = 0#

that is for #absq < 1#