# Determin whether the following geometric series converges or diverges. if converges finds it's sum Sum(upper^infinity,lower n=0 (-3)^n/5^(n+1)

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### 2 Answers

You need to perform the ratio test such that:

`lim_(n->oo) |(a_(n+1))/(a_n)| = lim_(n->oo) (((-3)^(n+1))/(5^(n+2)))/(((-3)^n)/(5^(n+1)))`

`lim_(n->oo) |(a_(n+1))/(a_n)| = lim_(n->oo) ((-3*(-3)^n)/((-3)^n))*((5*5^n)/(5^2*5^n))`

`lim_(n->oo) |(a_(n+1))/(a_n)| = (-3/5)*lim_(n->oo)(-3/5)^n`

Since `-3 < 5 => lim_(n->oo)(-3/5)^n = 0`

`lim_(n->oo) |(a_(n+1))/(a_n)| = (-3/5)*0 = 0`

Since performing the ratio test yields `lim_(n->oo) |(a_(n+1))/(a_n)| = 0 < 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=0)^oo ((-3)^n)/(5^(n+1)) = (1/5)sum_(n=0)^oo (-3/5)^n`

`sum_(n=0)^oo ((-3)^n)/(5^(n+1)) = (1/5)*(1/(1 - (-3/5)))`

`sum_(n=0)^oo ((-3)^n)/(5^(n+1)) = 1/(5 + 3)`

`sum_(n=0)^oo ((-3)^n)/(5^(n+1)) = 1/8`

**Hence, testing the convergence of the given series yields that the series converges absolutely and evaluating the sum yields `sum_(n=0)^oo ((-3)^n)/(5^(n+1)) = 1/8` .**

**Sources:**

`sum_(n=0)^oo (-3)^n/5^(n+1)`

To determine if the geometric series above converges or diverges, we need to convert it in a form

`sum_(n=0)^oo ar^n`

where r is the common ratio. When,

|r|<1 the series converges and if

|r|>1 the series diverges.

To do so, use the rule of exponents.

`sum_(n=0)^oo (-3)^n/5^(n+1) =sum_(n=0)^oo (-3)^n/(5^1*5^n)= sum_(n=0)^oo 1/5*(-3)^n/(5^n)= sum_(n=0)^oo 1/5(-3/5)^n`

So we have,

a= 1/5, r=-3/5 and |r|=3/5

*Since value of |r| is less than 1, the series converges.*

To determine the value of the series, use the formula:

`sum_(n=0)^oo ar^n= a/(1-r)`

Substittue the value of a and r.

`sum_(n=0)^oo (-3)^n/5^(n+1)= sum_(n=0)^oo 1/5(-3/5)^n = (1/5)/(1-(3/5))=(1/5)/(5/5+3/5)=(1/5)/(8/5)=1/8 `

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**Hence, the geometric series converges and its value is 1/8.**