SOLUTION: Find the sum of the series
$$1 + \frac{1}{2} + \frac{1}{10} + \frac{1}{20} + \frac{1}{100} + \cdots,$$
where we alternately multiply by $\frac 12$ and $\frac 15$ to get successiv
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Question 1209795: Find the sum of the series
$$1 + \frac{1}{2} + \frac{1}{10} + \frac{1}{20} + \frac{1}{100} + \cdots,$$
where we alternately multiply by $\frac 12$ and $\frac 15$ to get successive terms.
Found 2 solutions by CPhill, greenestamps:
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let the given series be $S$. We can write the series as:
$$S = 1 + \frac{1}{2} + \frac{1}{10} + \frac{1}{20} + \frac{1}{100} + \cdots$$
The terms can be written as:
$$S = 1 + \frac{1}{2} + \frac{1}{2 \cdot 5} + \frac{1}{2 \cdot 5 \cdot 2} + \frac{1}{2 \cdot 5 \cdot 2 \cdot 5} + \cdots$$
$$S = 1 + \frac{1}{2} + \frac{1}{2 \cdot 5} + \frac{1}{2^2 \cdot 5} + \frac{1}{2^2 \cdot 5^2} + \cdots$$
We can separate the series into two geometric series:
$$S = \left( 1 + \frac{1}{10} + \frac{1}{100} + \cdots \right) + \left( \frac{1}{2} + \frac{1}{20} + \frac{1}{200} + \cdots \right)$$
The first series is:
$$S_1 = 1 + \frac{1}{10} + \frac{1}{100} + \cdots = \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n$$
This is a geometric series with first term $a = 1$ and common ratio $r = \frac{1}{10}$. Since $|r| < 1$, the sum is:
$$S_1 = \frac{a}{1-r} = \frac{1}{1 - \frac{1}{10}} = \frac{1}{\frac{9}{10}} = \frac{10}{9}$$
The second series is:
$$S_2 = \frac{1}{2} + \frac{1}{20} + \frac{1}{200} + \cdots = \frac{1}{2} \left( 1 + \frac{1}{10} + \frac{1}{100} + \cdots \right)$$
$$S_2 = \frac{1}{2} \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n = \frac{1}{2} \cdot \frac{1}{1 - \frac{1}{10}} = \frac{1}{2} \cdot \frac{10}{9} = \frac{5}{9}$$
Therefore, the sum of the series is:
$$S = S_1 + S_2 = \frac{10}{9} + \frac{5}{9} = \frac{15}{9} = \frac{5}{3}$$
$$S = \frac{5}{3} = 1.666666\cdots$$
Final Answer: The final answer is $\boxed{5/3}$
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
+...
Split the series into two purely geometric series:
+...
and
+...
The infinite sum of the first series is
Note the second series is just half of the first, so the sum of the second series is 5/9.
The sum of the original series is then 10/9 + 5/9 = 15/9 = 5/3.
ANSWER: 5/3
Alternatively, we could group the terms in pairs to obtain a single purely geometric series.
The given series is then equivalent to the series
+...
The sum of that series is
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