Is sum_"n=1"^infty 3^(1 - n) * 5^(n/2) convergent or divergent? if it's convergent, find its sum.

1 Answer
Apr 12, 2018

sum_(n=0)^oosqrt(5)(sqrt5/3)^n=(3sqrt5)/(3-sqrt5) (Converges)

Explanation:

This very much resembles a geometric series; however, some simplification is needed:

a_n=3^(1-n)*5^(1/2n)

a_n=3/3^n*(5^(1/2))^n

a_n=3/3^n*sqrt(5)^n

a_n=3(sqrt5/3)^n

Thus, we have the series

sum_(n=1)^oo3(sqrt5/3)^n.

We see the common ratio, r=sqrt(5)/3approx0.74<1, so the series is convergent.

However, to find its value, we'll have to start at n=0, resulting in adding 1 to all exponents.

sum_(n=0)^oo3(sqrt5/3)^(n+1).

sum_(n=0)^oosqrt(5)/cancel3 cancel(3)(sqrt5/3)^n.

sum_(n=0)^oosqrt5(sqrt5/3)^n.

Now, recall that if |r|<1, sum_(n=0)^ooa(r)^n=a/(1-r).

With our rewritten series, a=5, r=sqrt(5)/3, so,

sum_(n=0)^oosqrt(5)(sqrt5/3)^n=sqrt5/(1-sqrt5/3)

sum_(n=0)^oosqrt(5)(sqrt5/3)^n=sqrt5/((3-sqrt5)/3)

sum_(n=0)^oosqrt(5)(sqrt5/3)^n=(3sqrt5)/(3-sqrt5)