# solve whether the given geometric series converges or diverges. if it converges, find its sum Sum(upper^infinity,lower n=1) 2^(n+1)/3^n-1

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### 1 Answer

Since the problem provides the information that the given series is a geometric series, it is recommended for you to perform the ratio test such that:

`lim_(n->oo)|a_(n+1)/a_n| = lim_(n->oo) |((2^(n+2))/(3^(n+1-1)))*((3^(n-1))/(2^(n+1)))|`

`lim_(n->oo) |a_(n+1)/a_n| = lim_(n->oo) |(4*2^n)/(3^n)*(3^n)/(6*2^n)|` = `4/6 = 2/3`

Since performing the ratio test yields `lim_(n->oo) |a_(n+1)/a_n| = 2/3 < 1` , hence, the series converges absolutely.

You need to evaluate the geometric series `sum_(n=0)^oo x^n = 1/(1-x)` , hence, reasoning by analogy, yields:

`sum_(n=1)^oo (2^(n+1))/(3^(n-1)) = sum_(n=1)^oo (2*2^n)/(3^n/3)`

`sum_(n=1)^oo (2^(n+1))/(3^(n-1)) = sum_(n=1)^oo 6(2/3)^n`

`sum_(n=1)^oo (2^(n+1))/(3^(n-1)) = 6*1/(1 - (2/3))`

`sum_(n=1)^oo (2^(n+1))/(3^(n-1)) = 3*6/(3-2)`

`sum_(n=1)^oo (2^(n+1))/(3^(n-1)) = 18`

**Hence, testing the convergence of the given series yields that the series converges absolutely and evaluating the sum yields `sum_(n=1)^oo (2^(n+1))/(3^(n-1)) = 18 .` **