Determin wheter the given geometric series converges or diverges. I it converges then find its sum Sum(upper^00,lower n=0) 13/(-5)^n

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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Since the given series is a geometric series, you need to perform the ratio test to tell if the series converges or it diverges such that:

`lim_(n->oo) |a_(n+1)/a_n| = lim_(n->oo) |13/((-5)^(n+1))*(-5)^n/13|` = `lim_(n->oo)|(-5)^n/((-5)^n*(-5))|` = `-1/5`

Notice that performing the ratio test yields `lim_(n->oo) |a_(n+1)/a_n| = -1/5 < 1` , hence, the series converges absolutely .

Since the given series is a geometric series, you need to remember that `sum_(n=0)^oo x^n = 1/(1 - x).`

`sum_(n=0)^oo 13/(-5)^n = 13 sum_(n=0)^oo (-1/5)^n`

`sum_(n=0)^oo 13/(-5)^n = 13*(1/(1 - (-1/5)))`

`sum_(n=0)^oo 13/(-5)^n = 13/(1+1/5)`

`sum_(n=0)^oo 13/(-5)^n = (13*5)/6`

`sum_(n=0)^oo 13/(-5)^n = 65/6`

Thus, testing the series yields that it converges absolutely and evaluating its sum yields `sum_(n=0)^oo 13/(-5)^n = 65/6` .

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lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

A geometric series has a form,

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

where r is the common ratio.

When |r| <1, the series is convergent. Otherwise, it is divergent.

To determine if the given geometric series is convergent or divergent, convert it to the form `ar^n` .

`sum_(n=0)^oo = 13/(-5)^n = sum_(n=0)^oo 13*(1/(-5)^n) = sum_(n=0)^oo 13*(-1/5)^n`

So we have,

`a=13`         `r = -1/5`         and           `|r|=1/5`

Since |r|<1, the series converges.

To determine the sum, use the formula of geometric series with infinite terms which is:

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

Substitute the values of a and r to the formula.

`sum_(n-0)^oo 13/(-5)^n = 13/(1-(-1/5))= 13/(5/5+1/5) = 65/6`

 Hence, the geometric series converges and its sum is  ` 65/6` .

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