Prove the following series converges and has the sum indicated using telescoping: $\sum^{\infty}_{n=1}\frac{2}{3^{n-1}}=3$

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I'm trying to compute the following limit for convergence using the telescoping property, however, I'm having difficulties simplifying the equation so the property can be used:

The following is the telescoping series: $\sum^{n}_{k=1}(b_k-b_{k+1}) = b_1-b_{n+1}$

The series I'm trying to telescope: $\sum^{\infty}_{n=1}\frac{2}{3^{n-1}}=3$

I have tried:

$\sum_{n=1}^{\infty}\frac{2}{3^{n-1}} \implies \frac{6}{3^n} = \frac{9}{3^n}-3^{n-1}$

When $n = 1$ then we get $\frac{9}{3^1}=3$?