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What is the sum of the below series?

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

here $$ r=\frac{1}{2} $$

from what I know about summations, when |r|<1, the sum is equal to:

$$ s=\frac{a}{1-r} $$

where a is the first member of the series. Here a equals zero, but the answer is 0.8. I don't know how to get to 0.8

Amirhosein
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    Perhaps rewriting the series may help... Something like $\frac 1{2^{n-1}}-\frac 1{6^{n-1}}$ to start. – abiessu May 22 '23 at 15:29
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    The important lesson to take away here is that formulas have hypotheses that need to be checked. The formula $\frac a{1-r}$ you mention is really the full implication "if a series is a geometric series with initial term $a$ and common ratio $r$ with $|r|<1$, then its value is $\frac a{1-r}$". One has to check that the series really is a geometric series before trying to use that fact; this series is not a geometric series. – Greg Martin May 23 '23 at 02:00

2 Answers2

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Notice that (as @abiessu pointed out) this is actually two summations in one, namely $$S=\sum_{k=1}^\infty\left(\frac1{2^{k-1}}-\frac1{6^{k-1}}\right)$$Re-indexing: $$S=\sum_{k=0}^\infty\left(\frac1{2^k}-\frac1{6^k}\right)$$We can split the sum since we know it converges. Since in both of the sums $a=1$, we have that $$S=\frac1{1-\frac12}-\frac1{1-\frac16}=2-\frac65=\frac45$$Which is equivalent to $\boxed{0.8}$

Kamal Saleh
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$$\sum_{n=1}^\infty \frac{3^{n-1}-1}{6^{n-1}}=\sum_{n=1}^\infty \frac{1}{2^{n-1}}-\sum_{n=1}^\infty \frac{1}{6^{n-1}}=\frac{1}{1-\frac12}-\frac{1}{1-\frac16}=\boxed{0.8}$$

Myst1cal
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