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)#
