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Which method allows me to calculate the value of the series as it tends to infinite?

$$\sum_{n=0}^\infty \frac{2^n}{3^n} $$

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    It is a geometric series – 2017-02-13
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    Rewrite it to $\sum (\frac 23)^n$. – 2017-02-13
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    Direct calculation of partial sums + taking the limit, ratio test, root test are the first three that come to mind. In all honesty, I wouldn't look for a fourth one. – 2017-02-13
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    The sum of series is $\dfrac{a_1}{1-r}$ where $a_1$ is the first term. – 2017-02-13

2 Answers 2

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Note that we know that If $|x|<1$, using a known result on Geometric Series, we have

$$\begin{align} \sum_{n=0}^\infty x^{n} &= \lim_{n\rightarrow\infty} \left(1 + x + x^2 + \cdots + x^n\right) \\ &= \lim_{n\rightarrow\infty} \frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}. \end{align}$$ Put $x=\frac{2}{3}$.

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A geometric series is defined by $\sum_{n=0}^{+\infty} r^n$.

If $|r| < 1$, the series is convergent and $\sum_{n=0}^{+\infty} r^n = \frac{1}{1 - r}$.

Rewrite you series $\sum_{n=0}^{+\infty} (\frac{2}{3})^n$. As $\frac{2}{3} < 1$, your series converges to $\frac{1}{1 - (\frac{2}{3})}$.