document.write( "Question 1209795: Find the sum of the series
\n" ); document.write( "$$1 + \frac{1}{2} + \frac{1}{10} + \frac{1}{20} + \frac{1}{100} + \cdots,$$
\n" ); document.write( "where we alternately multiply by $\frac 12$ and $\frac 15$ to get successive terms.
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Algebra.Com's Answer #850493 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "\"1%2B1%2F2%2B1%2F10%2B1%2F20%2B1%2F100%2B1%2F200\"+...

\n" ); document.write( "Split the series into two purely geometric series:

\n" ); document.write( "\"1%2B1%2F10%2B1%2F100\"+...

\n" ); document.write( "and

\n" ); document.write( "\"1%2F2%2B1%2F20%2B1%2F200\"+...

\n" ); document.write( "The infinite sum of the first series is

\n" ); document.write( "\"1%2F%281-1%2F10%29=1%2F%289%2F10%29=10%2F9\"

\n" ); document.write( "Note the second series is just half of the first, so the sum of the second series is 5/9.

\n" ); document.write( "The sum of the original series is then 10/9 + 5/9 = 15/9 = 5/3.

\n" ); document.write( "ANSWER: 5/3

\n" ); document.write( "Alternatively, we could group the terms in pairs to obtain a single purely geometric series.

\n" ); document.write( "\"1%2B1%2F2=3%2F2\"
\n" ); document.write( "\"1%2F10%2B1%2F20=3%2F20\"
\n" ); document.write( "\"1%2F100%2B1%2F200%2B3%2F200\"

\n" ); document.write( "The given series is then equivalent to the series

\n" ); document.write( "\"3%2F2%2B3%2F20%2B3%2F200\"+...

\n" ); document.write( "The sum of that series is

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