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$S_n = \sum_{k=1}^n { \frac{2}{3^k}}$

I am confused. Can I turn it into a limit approaching infinity even though $n$ is there? There was a previous question that asked me to find the first four terms of the sequence so does that mean $n = 4$? I need help please.

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    First, what is the question? – 2017-02-18
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    Does the sequence of partial sums converge or diverge? – 2017-02-18
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    You are restating the question, but what does it **mean**. Define **partial sums** for example – 2017-02-18

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One may recall the geometric series result (see here also) $$ \sum_{k=1}^nx^k=x \cdot\frac{1-x^{n}}{1-x}, \quad x \neq1, $$ giving $$ S_n = 2\sum_{k=1}^n { \frac{1}{3^k}}=\frac{2}{3} \cdot\frac{1-1/3^{n}}{1-1/3} $$ then, as $n \to \infty$, since $1/3^n \to 0$, one gets $$ S_n \to \frac{2}{3} \cdot\frac{1}{1-1/3}=1. $$

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    Goodness, I'm trying to teach someone XD – 2017-02-18