[superprismatic]

Suppose we allow digits in our base ten system to be any integer. Let's write the

"unusual" digits inside brackets, {}. Then 123 would have its usual meaning, but

1{-2}3 would be interpreted as 1×100 + (-2)×10 + 3 = 83, and 7{12}9 would be

interpreted as 7×100 + 12×10 + 9 = 829. Allowing these unusual digits, what would be

the limit of this series: 9/1, 98/12, 987/123, 9876/1234, 98765/12345, 987654/123456,

9876543/1234567, 98765432/12345678, 987654321/123456789, 9876543210/123456789{10},

9876543210{-1}/123456789{10}{11}, 9876543210{-1}{-2}/123456789{10}{11}{12}, ... ?

Please provide a proof of your answer.

[benjer3]

I'm assuming that you want the limit of the sequence. I don't think I've learned enough to do the whole proof, but I got an answer:

I made sequences for the numerator

***a_n = 10 a_{n - 1} - n + 9, where a₀ = 9

and the denominator

***b_n = 10 b_{n - 1} + n, where b₀ = 1

and plugged them into a graphing calculator. When I examined the ratio as n approaches infinity, I got

***8.79 or 800/91.

Edited March 16, 2011 by benjer3

[benjer3]

Sorry, made a silly mistake....

I made sequences for the numerator

***a_n = 10 a_{n - 1} - n + 10, where a₁ = 9

and the denominator

***b_n = 10 b_{n - 1} + n, where b₁ = 1

and plugged them into a graphing calculator. When I examined the ratio as n approaches infinity, I got 8.

[superprismatic]

Sorry, made a silly mistake....

I made sequences for the numerator

***a_n = 10 a_{n - 1} - n + 10, where a₁ = 9

and the denominator

***b_n = 10 b_{n - 1} + n, where b₁ = 1

and plugged them into a graphing calculator. When I examined the ratio as n approaches infinity, I got 8.

Thanks for correcting my "series" instead of "sequence" error. It's a

common mistake with me.

Yes, you got the correct answer. However, the meat of this problem lies

in the proof. Can you come up with one?

[EventHorizon]

First, put a decimal point in front of each one.

Notice that the sum of the numerator and the denominator approaches 1.1111111..... = 10/9

Let x = what the denominator approaches.

So the sequence approaches ((10/9)-x)/x = (10/9)*(1/x) - 1.

Lets take a closer look at x.

Subtract 1/9=.1111111111..... from x leaves .012345678...... = x/10.

x-(1/9) = x/10

(9/10) * x = 1/9

x = 10/81

So the sequence approaches (10/9)*(81/10) - 1 = 9-1 = 8.

Edited March 16, 2011 by EventHorizon

[superprismatic]

First, put a decimal point in front of each one.

Notice that the sum of the numerator and the denominator approaches 1.1111111..... = 10/9

Let x = what the denominator approaches.

So the sequence approaches ((10/9)-x)/x = (10/9)*(1/x) - 1.

Lets take a closer look at x.

Subtract 1/9=.1111111111..... from x leaves .012345678...... = x/10.

x-(1/9) = x/10

(9/10) * x = 1/9

x = 10/81

So the sequence approaches (10/9)*(81/10) - 1 = 9-1 = 8.

Looks good to me! You've got it!

[benjer3]

Very nice, Horizon. Simple and sweet. I was thinking too much into limits of geometric series and the like.