Number Of Digits In Powers Of Two

Synopsis : We use the properties of a logarithm to figure the general formula for $a_k$ . We then note that the limit of the expression consists of a ratio of two functions, the latter function is a strictly monotone and divergent sequence. So we can apply Stolz–Cesàro theorem to compute a simpler limit in hopes that if this new limit exists then so does, the limit in question and they are both equal in value. We then evaluate this simpler limit by accompanying a simple theorem: Squeeze theorem .

---

Recall that by properties of logarithms , we can compute the number of digits of number (written in decimal representation) $m$ to be $\lfloor \log_{10} m + 1 \rfloor$ . So in this case, we have $a_k =\left \lfloor \log_{10} 2^k + 1\right \rfloor = \lfloor k\log_{10} 2 + 1 \rfloor$ .

And we want to evaluate the limit, $\displaystyle \lim_{n\to\infty}\dfrac{a_1 + a_2 + \cdots + a_n}{n^2}$ .

Because the expression the denominator in the fraction is $n^2$ , which is a strictly monotone and divergent sequence, we can apply Stolz–Cesàro theorem , which states that

Let $(b_n)_{n\geq1}$ and $(c_n)_{n\geq1}$ be two sequences of real numbers. Assume that $(c_n)_{n\geq1}$ is a strictly monotone and divergent sequence and the following limt exists:

$\displaystyle \lim_{n\to\infty}\dfrac{b_{n+1} - b_n}{c_{n+1} - c_n} = \ell$ , then the limit $\displaystyle \lim_{n\to\infty} \dfrac{b_n}{c_n}$ also exists and it is equal to $\ell$ .

In this case, we let $b_{n} = a_1 + a_2 + \cdots + a_n \quad \text{ and } \quad c_n = n^2.$ Then $b_{n+1} - b_n = a_{n+1} = \lfloor (n+1) \log_{10}2 + 1 \rfloor \quad \text{ and } \quad c_{n+1} - c_n = 2n + 1 .$

Thus, $\displaystyle \lim_{n\to\infty} \dfrac{b_n}{c_n} = \lim_{n\to\infty} \dfrac{\lfloor (n+1) \log_{10}2 + 1 \rfloor}{2n+1}$ .

Because $z -1 < \lfloor z \rfloor \leq z$ . We have $(n+1) \log_{10}2 < \lfloor (n+1) \log_{10}2 + 1 \rfloor \leq (n+1) \log_{10}2 + 1$ , and so

$\displaystyle \lim_{n\to\infty}\dfrac{ (n+1) \log_{10}2 }{ 2n +1} < \lim_{n\to\infty} \dfrac{b_n}{c_n} \leq \lim_{n\to\infty} \dfrac{ (n+1) \log_{10}2 + 1}{2n+1}$

Because both the limits in left hand side of the inequality and the right hand side of the inequality evaluates to $\lim_{n\to\infty} \dfrac{ \left( 1 + \frac1n \right) \log_{10} 2}{2 + \frac1n} = \dfrac{(1+0) \log_{10} 2 }{2 + 0} = \dfrac12 \log_{10}2$

which is a finite value, then by Squeeze theorem, the limit in question also exists and is equal to $\dfrac12 \log_{10}2 = \log_{10} 2^{1/2} = \boxed{\log_{10} \sqrt2}$ as well.

Let $c = \log_{10} 2$ . Observe that $a_n = \lfloor \log_{10} 2^n \rfloor +1 = \lfloor cn \rfloor +1 = cn + O(1)$ . Now, the limit follows easily: $\lim_{n \rightarrow \infty} \frac{1}{n^2} \sum_{k = 1}^n a_k = \lim_{n \rightarrow \infty} \frac{1}{n^2} \sum_{k = 1}^n cn + O(1) = \lim_{n \rightarrow \infty} \frac{c(n+1)}{2n} + O(1/n) = c/2 = \log_{10} \sqrt{2}.$