Define a sequence of numbers like this: x1=π;xn+1=xn−(the nth digit of π)∗10−n[math]x_1=\pi; x_{n+1}=x_n-(\mbox{the nth digit of } \pi)*10^{-n}. Here the 1st digit of π[math]\pi is 1, the second 4, etc. Is this sequence convergent? If it is convergent, what is its limit? Give your answer no or enter the limit.

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BlackTom AI · Accepted Answer

The problem defines a sequence by x1 = π and x_{n+1} = x_n − (the nth digit of π) × 10^{−n}, where the nth digit of π refers to the decimal digits 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, … in order.

To understand convergence, express x_{N+1} in terms of x_1 by unrolling the recurrence:
- x_2 = x_1 − d_1·10^{−1}
- x_3 = x_2 − d_2·10^{−2} = x_1 − d_1·10^{−1} − d_2·10^{−2}
- x_4 = x_3 − d_3·10^{−3} = x_1 − d_1·10^{−1} − d_2·10^{−2} − d_3·10^{−3}
and so on, where d_n denotes the nth digit of π.
Hence, after N steps,
x_{N+1} = x_1 − ∑_{n=1}^N d_n · 10^{−n}.
If the limit x_n as n → ∞ exists, call it L. Then L = x_1 − ∑_{n=1}^∞ d_n · 10^{−n}.
Now, note that the digits d_n are the decimal digits of π, so the sum ∑_{n=1}^∞ d_n · 10^{−n} equals 0.d_1d_2d_3… in base 10, which is the decimal 0.14159265358979323846…
This trailing decimal is precisely π − 3, because π = 3.141592653589793…
Therefore, L = π − (π − 3) = 3.

Thus, the sequence does converge, and the limit is the real number 3.0 (i.e., 3).

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