![\"\"](\"/images/stars/stars-12.gif\")
You can put this solution on YOUR website!
Let the given series be $S$. We can write the series as:
\n" ); document.write( "$$S = 1 + \frac{1}{2} + \frac{1}{10} + \frac{1}{20} + \frac{1}{100} + \cdots$$
\n" ); document.write( "The terms can be written as:
\n" ); document.write( "$$S = 1 + \frac{1}{2} + \frac{1}{2 \cdot 5} + \frac{1}{2 \cdot 5 \cdot 2} + \frac{1}{2 \cdot 5 \cdot 2 \cdot 5} + \cdots$$
\n" ); document.write( "$$S = 1 + \frac{1}{2} + \frac{1}{2 \cdot 5} + \frac{1}{2^2 \cdot 5} + \frac{1}{2^2 \cdot 5^2} + \cdots$$
\n" ); document.write( "We can separate the series into two geometric series:
\n" ); document.write( "$$S = \left( 1 + \frac{1}{10} + \frac{1}{100} + \cdots \right) + \left( \frac{1}{2} + \frac{1}{20} + \frac{1}{200} + \cdots \right)$$
\n" ); document.write( "The first series is:
\n" ); document.write( "$$S_1 = 1 + \frac{1}{10} + \frac{1}{100} + \cdots = \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n$$
\n" ); document.write( "This is a geometric series with first term $a = 1$ and common ratio $r = \frac{1}{10}$. Since $|r| < 1$, the sum is:
\n" ); document.write( "$$S_1 = \frac{a}{1-r} = \frac{1}{1 - \frac{1}{10}} = \frac{1}{\frac{9}{10}} = \frac{10}{9}$$
\n" ); document.write( "The second series is:
\n" ); document.write( "$$S_2 = \frac{1}{2} + \frac{1}{20} + \frac{1}{200} + \cdots = \frac{1}{2} \left( 1 + \frac{1}{10} + \frac{1}{100} + \cdots \right)$$
\n" ); document.write( "$$S_2 = \frac{1}{2} \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n = \frac{1}{2} \cdot \frac{1}{1 - \frac{1}{10}} = \frac{1}{2} \cdot \frac{10}{9} = \frac{5}{9}$$
\n" ); document.write( "Therefore, the sum of the series is:
\n" ); document.write( "$$S = S_1 + S_2 = \frac{10}{9} + \frac{5}{9} = \frac{15}{9} = \frac{5}{3}$$
\n" ); document.write( "$$S = \frac{5}{3} = 1.666666\cdots$$\r
\n" ); document.write( "\n" ); document.write( "Final Answer: The final answer is $\boxed{5/3}$
\n" ); document.write( " \n" ); document.write( "