Why does 98765.../12345... converge to 8?

Question

The terms in the sequence 9/1, 98/12, 987/123, 9876/1234,... approach 8 as the digits in the numerator and the denominator approach infinity. Why is this?

The terms in the sequence 9/1, 98/12, 987/123, 9876/1234,... approach 8 as the digits in the numerator and the denominator approach infinity. Why is this?

Clarification: The numerators and denominators beyond (987654321, 123456789) are (9876543210, 1234567900), (98765432099,12345679011),...

In other words, where n is the number of digits, for the numerator, f(n)=f(n-1)*10+(10-n), and for the denominator, g(n)=g(n-1)*10+(n).

Answer 1

Okay, the terms in your sequence may be represented as the following quotient of two series:

[k=1, n]∑((10-k)*10^(n-k)) / [k=1, n]∑(k*10^(n-k))

It turns out that this is equivalent to the simpler formula:

9 * (10^(n+1) - 10) / (10^(n+1) - 9n - 10) - 1

It is then easy to see that the limit as n approaches infinity of this expression is 8, since:

[n→∞]lim 9 * (10^(n+1) - 10) / (10^(n+1) - 9n - 10) - 1
= [n→∞]lim 9 * (1 - 10^(-n)) / (1 - 9n/10^(n+1) - 10^(-n)) - 1
= 9 * (1-0)/(1 - 0 - 0) - 1
= 9 - 1
= 8

Thus all that remains is to prove the assertion that the two expressions are in fact equivalent. I should warn you that the proof of this is _really ugly_. But if you have the stamina to slog through it, here it is. Recall the original expression:

[k=1, n]∑((10-k)*10^(n-k)) / [k=1, n]∑(k*10^(n-k))

Let us start by breaking this into two fractions:

[k=1, n]∑(10^(n+1-k)) / [k=1, n]∑(k*10^(n-k)) - [k=1, n]∑(k*10^(n-k)) / [k=1, n]∑(k*10^(n-k))

Greatly simplifying the fraction on the right:

[k=1, n]∑(10^(n+1-k)) / [k=1, n]∑(k*10^(n-k)) - 1

Now, let's start with the numerator of the fraction on the left. This can be represented with the formula for the sum of a geometric series:

(10^n - 1)/(1-1/10) / [k=1, n]∑(k*10^(n-k)) - 1

Simplifying:

(10^(n+1) - 10)/9 / [k=1, n]∑(k*10^(n-k)) - 1

To handle the series in the denominator, first we factor out 10^(n+1):

(10^(n+1) - 10)/9 / (10^(n+1) [k=1, n]∑(k*10^(-k-1))) - 1

Now, we employ a trick. We represent the series as a function of a dummy variable p which we evaluate at 10. This gives us:

(10^(n+1) - 10)/9 / (10^(n+1) {[k=1, n]∑(k*p^(-k-1)) @ p=10}) - 1

Of course, this is immediately recognized as the following derivative:

(10^(n+1) - 10)/9 / (10^(n+1) {d/dp [k=1, n]∑(-p^(-k)) @ p=10}) - 1

Evaluating the terms now using the formula for the sum of a geometric series:

(10^(n+1) - 10)/9 / (10^(n+1) {d/dp (-p^(-1) + p^(-n-1))/(1-1/p) @ p=10}) - 1

Actually taking the derivative:

(10^(n+1) - 10)/9 / (10^(n+1) {((p^(-2) - (n+1)p^(-n-2)) * (1-1/p) - (-p^(-3) + p^(-n-3)))/(1-1/p)² @ p=10}) - 1

Evaluating the function at p=10:

(10^(n+1) - 10)/9 / (10^(n+1) ((1/100 - (n+1)10^(-n-2)) * (1-1/10) - (-10^(-3) + 10^(-n-3)))/(1-1/10)²) - 1

Now that we have a closed-form expression, we attempt to simplify it:

(10^(n+1) - 10)/9 / (10^(n+1) ((1/100 - (n+1)10^(-n-2)) * 9/10 + 10^(-3) - 10^(-n-3))/(9/10)²) - 1

Further (handling the 10² in the denominator of the denominator of the denominator):

(10^(n+1) - 10)/9 / (10^(n+1) ((1 - (n+1)10^(-n)) * 9/10 + 10^(-1) - 10^(-n-1))/9²) - 1

And simplifying further (handling the 9² in the denominator of the denominator):

9 * (10^(n+1) - 10) / (10^(n+1) ((1 - (n+1)10^(-n)) * 9/10 + 1/10 - 10^(-n-1))) - 1

And further (distributing the 9/10):

9 * (10^(n+1) - 10) / (10^(n+1) (9/10 - 9*(n+1)10^(-n-1) + 1/10 - 10^(-n-1))) - 1

And still further (distributing 10^(n+1)):

9 * (10^(n+1) - 10) / (9*10^n - 9*(n+1) + 10^n - 1) - 1

The function is beginning to look somewhat manageable now. We make another simplification:

9 * (10^(n+1) - 10) / (10^(n+1) - 9n - 10) - 1

And we are done.

Answer 2

Is this just a question for curiosity? Or, are you asking for some rigorous, technical mathematical explanation? If you are just curious, the basic explanation is that your expression is based on a progression the involves multiplying the numerator by 10 and subtracting a small number from it while multiplying the denominator by 10 and adding a small number to it. For example, the first expression numerator is 10^1 - 1 or 9 while the denominator is 10^0. The second expression in the sequence is 10^2 - 2 or 98 while the denominator is 10^1 + 2 or 12. At this point, the expression already evaluates to 8.2 which is close to 8. Then, as you move progressively through the sequence you keep increasing both the numerator and denominator each another power of 10. But, as another factor in arriving at the next numerator you are also SUBTRACTING a relatively small number while you are also ADDING a relatively small number to the denominator calculation. This means the numerator is growing at a rate a little LESS than one power of 10 while the denominator is increasing at a rate a little MORE than on power of 10. So, when the numerator grows a little slower than the denominator grows the resulting fraction gets smaller over time. This difference in the rate of change between the numerator and denominator is what drives the expression closer and closer to 8 as you go through the sequence.

Answer 3

You can figure out a series expansion for these terms.

For example consider the Maclaurin series of
f(x) = 1/(1-x) = 1 + x + x^2 + x^3 + .....
Now differentiate this to get
f'(x) = 1 / (1-x)^2
= 1 + 2x + 3x^2 + 4x^3 + ...

f'(0.1) = 1 + 0.2 + 0.03 + ... = 1.23456789...
= 1 / (1 - 0.1)^2 = 100 / 81

If you multiply that by 8, you get 800 / 81
= 9.87654321...

Answer 4

You are not being clear, what pattern do you use after you reach 987 654 321 / 123 456 789 ??

Answer 5

u make my brain hurt