Series

3^n/9^(2n + 1) = 1/3^(3n + 2) = (1/9)(1/3^(3n))

So, by ratio test, the series converges

Σ 3^n/9^(2n + 1)

= (1/9) Σ (1/3^(3n))

= (1/9)(1 + 1/3^3 + 1/3^9 + ...)

= (1/9)(1/(1 - 1/27))

= (1/9)(27/26)

= 3/26

Hope it helps

i forgot to add n=0.
sorry about that

Tony, the first term is for n = 0, so the first term is 1/9.

Therefore, the given answer of 3/26 is actually correct.

(1/9) / ( 1 - 1/27 ) = 3 / (27 - 1) = 3 / 26

2014-04-20 23:20:00 補充：
Also, the general term in a geometric series is (a rⁿ⁻¹) or equivalent.

Therefore,
3ⁿ / 9²ⁿ⁺¹ = (3/9²)ⁿ / 9 = (1/9) (1/27)ⁿ is clearly a member of it.

Just figure out a number raise to the power of n, and look into it and check whether it has an absolute value less than 1.

Done.

2014-04-20 23:20:50 補充：
BTW, you can also check that 1/234 + 1/9 = 3/26.

2014-04-21 17:52:33 補充：
No worry~

You have a good heart to help the poster as well.