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anonymous
 5 years ago
The meaning of the decimal representation of a number 0.d1d2d3 . . . (where the digit i is one of the numbers 0, 1, 2, . . ., 9) is that 0.d1d2d3d4 . . . = d1/10 + d2/10^2 + d3/10^3 + d4/10^4 + . . . Show that this series always converges.
anonymous
 5 years ago
The meaning of the decimal representation of a number 0.d1d2d3 . . . (where the digit i is one of the numbers 0, 1, 2, . . ., 9) is that 0.d1d2d3d4 . . . = d1/10 + d2/10^2 + d3/10^3 + d4/10^4 + . . . Show that this series always converges.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Geeze, what grade is this?? ((I do not know the answer.))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is calculus 2, so its gonna be tricky!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1so you are having trouble this converges: sum(1/10^i,i=1..n)? if this sum coverges then sum(di/10^i,i=1..n) should converges i think

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1ratio test we can show it converges absolutely lim d{n+1}/10^(n+1)/d{n}/10^n n>inf = " d{n+1}/10^n*10*10^n/d{n} = " 1/10*d{n+1}/d{n} the digits can range form 0 to 9 the biggest number that can be in that absolute value thing is 9/1 so 1/10*9=9/10<1 so it converges absolutely

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you! this helps immensely. I had a solution worked out involving the "r" value being less than one for 9/10^n (geometric series), making it convergent
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