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)