Sum of infinite series....?

If n starts from 1, then the sum is (2/5)/[1-(6/25)] = 10/19

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Ideas to get r: r = [(2/5)^(n+1)][(3/5)^(n)] / [(2/5)^n][(3/5)^(n-1)] = (2/5)(3/5) = 6/25

It can be written as a geometric series, a sum of terms of the form a*r^n.

[(2/5)^n][(3/5)^(n-1)] = (2/5)^n (3/5)^n (3/5)^-1

= (6/25)^n * (5/3) so a = 5/3 and r = 6/25

You didn't say what your starting term was, but the sum of a*r^n for n = 1, 2, ... is a*r/(1 - r).

The sum from n = 0 to infinity is a/(1 - r).

I assume you find the sum from n= 1 to infinity.

No, you cannot find the sum of each series and multiply them.

Your series is of the form a^n * [ b^ (n-1) ] = a^n * ( b^n /b ) = [ (ab)^n /b ]

So use the formula of the geometric series.