Understanding the Sum of Digits Function

To prove that the sum of the digits of (2^n) in base 10 tends to infinity as (n) increases, we can employ a methodical approach. This involves analyzing the behavior of the sum of digits as (n) grows. Let (S_n s(2^n)) denote the sum of the digits of the number (2^n) in base 10. Our objective is to show that (S_n rightarrow infty) as (n rightarrow infty).

Bounding (S_n)

The number (2^n) typically has approximately (n log_{10}2) digits. This can be estimated using:

[d approx lfloor n log_{10}2 rfloor 1.]

Here, the number of digits (d) in (2^n) grows linearly with (n). This insight is crucial because it sets the stage for understanding the behavior of (S_n) as (n) increases.

Upper Bound for the Sum of Digits

Each digit in a number can take a value from 0 to 9. Therefore, the maximum possible sum of the digits (S_n) can be crudely bounded by:

[S_n leq 9 cdot d approx 9 cdot n log_{10}2 1 approx 9n log_{10}2.]

This upper bound indicates that (S_n) can grow linearly, but let's explore a more robust lower bound.

Lower Bound for the Sum of Digits

To show that (S_n) tends to infinity, we need a lower bound. Let's consider the fact that as (n) increases, (2^n) grows exponentially. For sufficiently large (n), the number (2^n) will have many non-zero digits.

Using Density Arguments:

The sum of the digits (S_n) can be seen as a function reflecting the distribution of values in the binary representation of (2^n) translated to decimal. As (n) increases, (2^n) is increasingly likely to have more non-zero digits because it covers a wider range of values in base 10. This suggests that the sum of digits will grow over time.

Formal Proof

Let (a) be any number with digits (a_k cdots a_0). Define a function (phi a_i) as the (i)th digit of (a). Extend (phi) to be defined over all (i in mathbb{R}) by making boxes, as illustrated in the diagram. For (2^n), this function (phi) will cycle through a set of values, such as (2, 4, 8, 6, 2, cdots).

Notice that the sum (S_a) is just the area under the boxes in our diagram:

[S_a int_{0}^{n} phi , dx.]

Since (phi) is always non-negative and ranges from 0 to 9, the integral, which is (S_a), must be non-decreasing. If (S_a) were constant, all the digits of (a) would eventually become 0, making (phi) eventually all zero. However, the last digit of (2^n) cycles through (2, 4, 8, 6, 2, cdots), so it cannot be all 0 after any point. Therefore, (S_{2^n}) must be increasing.

The formalization involves nudging (phi) slightly to make it a proper function, but the integral remains unchanged. This method provides a rigorous proof that (S_n) increases as (n) grows.

Conclusion

Since the number of digits in (2^n) increases with (n) and the maximum sum of the digits also increases, we can conclude:

[S_n rightarrow infty text{ as } n rightarrow infty.]

This result demonstrates that the sum of the digits of (2^n) in base 10 tends to infinity as (n) increases.