jasperE3 wrote: Let $\alpha_a$ denote the greatest odd divisor of a natural number $a$, and let $S_b=\sum_{a=1}^b\frac{\alpha_a}a$ Prove that the sequence $S_b/b$ has a finite limit when $b\to\infty$, and find this limit. We notice that for each $a \in \mathbf{N}$ one has $\alpha_a \le a$ and thus $S_b \le b \implies \frac{S_b}{b} \le 1$ for all $b \in \mathbf{N}.$ Now fix some odd positive integer $t$ and let $A(t)$ denote the set of all and only those positive integers $k$ for which $\alpha_k = t,$ that is $A(t) := \{ k \in \mathbf{N} ~ : ~ \alpha_k = t \}$ defined for each odd positive integer $t.$ What can we say about the elements in $A(t)$? Well, notice that a positive integer $g$ has $\alpha_g = q$ for an odd positive integer $q$ if and only if $q| g$ and $g/q$ is not divisible by any odd number other than $1$ (this is because if an odd positive integer $h| g/q$ and $h \ne 1$ then $hq$ would have been an odd divisor of $g$ and here the condition $h \ne 1$ would say that $hq > q).$ Thus, $A(t) = \{ t \cdot 2^k ~ : ~ k \in \mathbf{N}_0 \}$ for each odd positive integer $t.$ Now, for each odd $t \in \mathbf{N}$ notice that $$\sum_{j \in A(t)} \frac{\alpha_j}{j} = \sum_{k=0}^{\infty} \frac{t}{t \cdot 2^k} = \sum_{k=0}^{\infty} \frac{1}{2^k} = 2.$$Also notice that the inequality $\alpha_y \le y$ for each positive integer $y$ grants us the impunity to write $$\sum_{a=1}^{b} \frac{\alpha_a}{a} \le \sum_{1 \le t \le b~ \& ~ t \text{ is odd } } \sum_{j \in A(t)} \frac{\alpha_j}{j},$$and thus by the preceding inequality one has $\sum_{a=1}^{b} \frac{\alpha_a}{a} \le$ twice the number of odd numbers in the interval $[1,b] \sim b.$ (Here $\sim$ denotes asymptotic equivalence or that the ratio LHS/RHS approaches the limit 1 as the variable ($b$ in this case) approaches infinity.) Thus, we obtain that the limit $\lim_{b \to \infty} \frac{S_b}{b} =1$ (thanks to the following which I leave as an exercise). Exercise: Suppose $(x_n)_{n=1}^{\infty}, (y_n)_{n=1}^{\infty}$ and $(z_n)_{n=1}^{\infty}$ be three sequences of real numbers such that $\bullet$ we have $x_n \le y_n$ for all positive integers $n > N_1$ for some positive integer $N_1$; $\bullet$ and also $y_n \le z_n$ for all positive integers $n > N_2$ for some natural number $N_2$; $\bullet$ and $\lim_{n \to \infty} \frac{x_n}{z_n} = 1.$ Then show that $\lim_{n \to \infty} \frac{x_n}{y_n} = 1.$