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+...#?

Question

Isabella Ackerman · Accepted Answer

undefined The given series is a geometric series with first term $ a = 1 $ and common ratio $ r = \frac{1}{3} $.

1. The nth partial sum of a geometric series is given by the formula: $ S_n = \frac{a(1 - r^n)}{1 - r} $.

2. To determine whether the series converges, recall the condition for convergence of a geometric series: It converges if and only if $ |r| < 1 $.

3. If $ |r| < 1 $, the sum of the geometric series is given by the formula for the sum of an infinite geometric series: $ S = \frac{a}{1 - r} $.

Thus, for the given series:

1. The nth partial sum is $ S_n = \frac{1(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}} $.

2. The series converges since $ |r| = \frac{1}{3} < 1 $.

3. The sum of the series is $ S = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} $.

Samantha Adkison · Answer

#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#